12VAC is the RMS value of output sine. The bridge rectifier allows you to achieve a much higher voltage though, because then the storage cap is charged to up the peak voltage of sine (which is SQRT(2)=1.414 times higher than RMS value) minus the voltage drop of two diodes.
So, you will idealy see a voltage at storage cap of about RMS_value x 1.414 - 1.5V = 12VAC x 1.414 - 1.5V = 15.5V.

But this simple formula is only valid for very strong transformers, which do not show relevant ohmic resistance of secondary winding, and if the nominal load is drawn from the transformer. But in reality the finite transformer impedance leads always to a lowering of output voltage of up to some volts under nominal load.

A special case are very tiny transfomers. These can show a very drastical difference between output voltage with and without load! I have seen transformers which produced up to 3 times more output voltage when running without load!! This mechanismn is very probably responsible for, that you see a voltage of 19V instead of "calculated" 15.5V.

How big should the storage cap be?
With a standard bridge rectifier the storage cap is filled every 10msec (50Hz mains). At worst case condition the storage cap must deliver current for 10msec. In this time period he loses voltage. How much?

C = Q / U = I x t / U

where C is capacitance of storage cap, I is the current drawn from it, U is the voltage it loses and t is the time period where the current is drawn.

This yields:

U = I x t / C

An example:
If C = 1000µF, t = 10msec and I = 100mA, then U = 1V

In practice the storage cap must be choosen to be big enough to deliver current without huge voltage loss. On the hand, the storage cap should not be choosen too high though, because the higher this cap, the less time the transfomer has to give it back the charge and the higher and more narrow the charge peak current is. This can result in enormous interference.

So, a 5000µF storage cap for a 100mA application is highly overdimensioned.

Kai