Assume that I have a 220 to 12 volts transformer with a nominal output current of 3 A (marked "12 V 3 A"). Now what does happen if I connect the two output wires to each other or to a load with very low impedance? Does the current exceed 3 A by a large margin? I don't think so. But is the secondary coil designed to withstand such currents? If the temperature gets too high, the insulation of the coil wires would be damaged, which is obviously dangerous.
If you short the output of a good/efficient transformer, the secondary current will be very high, and the transformer will likely overheat and be damaged.
However, there are intentionally lossy ("impedance protected") transformers that will limit the load current to a safe value to prevent damage to the transformer in case of short circuits or overloads (often used to power doorbells and similar applications).
There's no upper limit to the current that will flow in that case. The transformer will cheerfully draw, up to its impedance and inductive limits, until it melts the secondary down into copper slag. Or, the thin primary wires will overheat and blow like a fuse.
We see it a lot where people think a component has built-in overcurrent protection because it'd be cool if it did. No; only fuses and breakers do. Most ratings are just the maximum safe load, and you are expected to design so your equipment cannot exceed it. That is something UL will want to see before they agree to list your equipment.
You can put L-rated (75 mph) tires on your Ferrari and drive it 200 mph, and you can do it all day. It's not a good idea, though.
An ideal transformer would give you infinite current and not heat up at all. But unfortunately, you don't have one of those. A real transformer has all kinds of parasitic effects that make it difficult to model with absolute accuracy, but the one that we're interested in now is the winding resistance.
Each winding has its own resistance that is simply the wire itself that it's made from. So you can find it easily with an ohmmeter. (not when it's plugged in, of course!) Now take that resistance in series with an ideal transformer, and you get both a crude voltage/current profile and a crude estimate of how much it'll heat up with a given load.
For example, if your secondary winding resistance is 1 ohm*, then the turns ratio might be fudged to produce 15V open-circuit, 12V @ 3A, and 15A short-circuit. The winding itself would dissipate 0W, 9W, and 225W, respectively.
* I'd be surprised if it was that high, but it makes the math easy. We'll just go with it for now.
Of course, there are other losses as well:
- The primary resistance heats up too.
- The iron core gets "massaged" by the magnetic field which causes molecular/atomic friction. So some of the energy heats the core instead of getting transferred.
- The magnetic field doesn't completely couple to the secondary but "spills" some to the surrounding area. So some of the energy couples into other circuitry that may be nearby, instead of getting transferred where it's supposed to.
- For low-power devices, a wall-wart is a convenient way to avoid this. Even large electronics, like analog audio mixers for example, may have an external power supply for exactly this reason.
But you get the basic idea.
You can set an upper bound on the current by measuring the DC resistance of the primary and secondary and then calculating:
Imax = 220 / (Rprimary + (220/Vout)^2 * Rsecondary)
(Vout here is the open circuit voltage as we are establishing the turns ratio N)
There is an efficient winding principle that tells you that the most efficient use of copper is when primary and secondary copper losses are equal. So this lets us estimate the secondary (as very low resistances can be hard to measure with the ohms range)
Imax = 220 / (2* Rprimary)
Why don't you try this, and report back to us. I suspect you will find Imax to be shockingly high.
Real current is lower due to leakage reactance, but this is useful because the current cannot be higher.