The first thing to understand is that an 'ideal' multimeter measuring current will have zero resistance across its terminals which means it will allow as much current to pass as the circuit it is in can provide. Such an uncontrolled flow can be a source of danger. In a real multimeter, however, this resistance will not actually be zero but it will be as close to zero as the designer can get - so it is always wise to think of your multimeter (in current mode) as being a simple piece of wire between between one probe and the other.
Now imagine what would happen if you stuck a piece of wire into the active and neutral of a mains power point that was switched on!!! Some fireworks, a blown fuse or both - as well as a risk of injury to yourself. This is the sort of problem that can happen when you don't use measuring equipment properly.
In the case of your wall wart, this is exactly the sort of thing you were doing - but you were fortunate in that things were on a smaller scale and the meter was designed to be able to carry that current. I would also guarantee you were overloading the wall wart and if you had left it connected it would have heated up very quickly and very likely have been destroyed.
To understand why you kept getting 2A, I will offer you the following
very simplified diagram. This represents your wall wart - but only in a very basic sense. It is missing a lot of detail that you would find in a real device, but it shows why you would get the same current reading. To make the maths simpler, I will use the 'ideal' current measuring meter that has zero ohms between the probes.
The story goes like this...
As shown, there are four sections of the transformer that generate 3V each - but the wire that is used in each section has a resistance of 1.5 ohms.
* When you select 3V, there is only 1 section used and it has a resistance of 1.5 Ω. Using Ohm's Law V=IR, you get a current of 2A.
* When you select 6V, there are now two sections used and while this gives you 6V, the resistance across the two sections is 3 Ω. Again, using Ohm's Law, you get a current of 2A.
* When you select 9V, there are now three sections used and while this gives you 9V, the resistance across the three sections is 4.5 Ω. Once more using Ohm's Law, you get a current of 2A.
* When you select 12V, there are now four sections used and while this gives you 12V, the resistance across the four sections is 6 Ω. Yet again using Ohm's Law, you get a current of 2A.
This resistance is buried inside the wall wart and something you can't really do anything about it, other than include it in your calculations. This sort of internal resistance has a name: "Equivalent Series Resistance" or ESR and it is found in every electrical and electronic device - unless it happens to be a superconductor.
It is this ESR and how it is distributed through the wall wart that gives rise to the "constant 2A" phenomenon you observed.
For those wanting to challenge the above, I would like to add that there are several ways of providing a multi-voltage wall wart - and the above is only showing
one of those in a
very basic conceptual form,
purely for the purposes of education,
not as a construction project.