Hi,
if you want to do it pyhsically and mathematically correct you need to know the values of C1, C2 and the impedance charging these Cs (this also includes the ESR of the capacitors)
It´s a lot of math, so better run a simulation.
***
A very inaccurate estimation:
positive side output current is: 1.3A + 0.15A = 1.45A DC
negative side output current is 0.05A + 0.15A = 0.2A DC
the voltage across the bulk capacitors is up to 15V x sqrt(2) (for ideal transformer and ideal mains supply, which is not realistic)
so lets say about 21V (rectifier loss also not included).
No we have about 21V ... and combined of both sides we have an average current of 1.65A.
Now charging of the capacitors (C1, C2) happens oly at the peaks of the sine mains voltage. let´s say twice in a period for maybe just 20% of time.
Again if we (unrealistically) say the current is like a square wave, then the current is 1.65A / 20% for 20% of the time.
So 8.25A for 20% of time.
For square wave currents: the RMS current is the peak current * sqrt(duty cycle), so 8.25A x sqrt(20%) = 8.25A X 0.5 = 4.125A RMS (combined for both sides)
Each output is rated with 2A RMS, making a total of 4A RMS.
--> with the given load the transformer is yet a bit overloaded.
While the load (output) power is just 1.65A x 15V = 24.75W
The transformer average output power may be about 35W (The difference of about 10W is dissipated in diodes and voltage regulators as heat)
But the apparent transformer output power is about 60VA due to power factor.
Klaus