Maximum resistance allowing for short circuits

Here is a redrawn circuit with a Lm321 IC used as the current limit detector. The output from the detector could also go to the gate of a SCR, with the Anode going to pin 1 of your regulator and the cathode tied to ground, killing the current regulator output. I designed and used this circuit in a cnc stepper motor driver circuit.

Use 0.1 ohm resistor at 3 Watts and 2.2k (1/4 to 1/2) watt. A 2N5060 is a sensitive gate low current SCR. The output Voltage may not go to zero volts but it will be close enough to Stop most current flow. You are right their is a minimum current requirement to keeping an SCR Latched. A simple test put an SCR on pin 1 Anode and Cathode to ground then trigger the gate. If it holds it will work. The gate will trigger just by touching it with your finger. Yes the SCR will stay latched until the 12 volts goes to Zero, or a momentary NC switch is installed between the Cathode and Ground. 2N5060 Low Holding Current = 5 mA Maximum Current Require to keep latched.

A transistor circuit will work if it is stabilized with a capacitor, values I am no sure of, That would have to be experimented with, otherwise the voltage will oscillate between high and low.

You can look at it this way (since you said you want an answer based on math):

The equation for the current through a capacitor is the following:

\$ I_C = C \dfrac{dV}{dt} \$

So at \$ t=0 \$, when the switch is closed, there's a voltage across the capacitor and \$ \frac{dV}{dt} \$ is really large (theoretically infinite) because there is a change in the voltage in technically 'no time,' therefore the slope \$ (\frac{dV}{dt}) \$ approaches infinity at \$ t=0 \$. Since \$ \frac{dV}{dt} \$ is in theory infinity so is the current through the capacitor (they are directly proportional). That explains why it looks as a short at time zero.

As you probably know differentiation does not exist at sharp edges and this is exactly what is happening from a mathematical point of view. Hope it helps!