Norton equivalent theorem
If there is a short-circuit between a and a' then all the current will take that route in preference to the parallel route with resistance.
Since your current source is pushing out 2 A then all of it will flow through the short-circuit so \$ i_N = 2 \ \mathrm A \$.




From comments:


  Why do all the current take the a-a' route in preference to parallel route with resistance?


Quick answer: because it's easier.

Longer answer: to get current to flow through a resistor you need some voltage across it. By definition you can't have voltage across a short circuit¹, therefore you won't have voltage across the resistor and can't push any current through it².



¹ A short-circuit has resistance = 0. Voltage across it is given by Ohm's Law, \$ V = IR = I \times 0 = 0 \$ no matter what the current is.

² Again, from Ohm's Law, \$ I = \frac {V}{R} = \frac {0}{R} = 0 \$ through any resistance in parallel with a short circuit.