Hi,
Recently, the 78xx series gives output current of 1A. 7812 cannot output 2A. 7805 cannot output 2A either. Maybe you are going to find a way of safely connecting two of your circuits in parallel.
Your transformer too must be properly rated. Make some rough calculation like this for your transformer:
The ON semiconductor datasheet gives "Output voltage max = 5.25V @ (5.0mA <=Io <=1.0A, Pd <=15W) 7.5Vdc <= Vin <= 20V." And "Pd <= 15W" for 7812.
Using worst case;
7805 Pin = 5.25x1 + 15 = 20.25W;
7812 Po = 7805 Pin + 12×1 = 20.25+12.25 = 32.5W.
7812 Pin = 32.5 + 15 = 47.5W.
This is for one part.
2 x 7812 Pin = 47.5×2 = 95W for both parts.
Add rectifier diodes loss to this 95W. If its a 24V transformer then 95/24 = 3.958 = 4A. The loss may be 1.1V at 4A = 4.4W for one diode and hence 8.8W for two diode. Pin before bridge stage = 8.8 + 95 = 103.8W. Your transformer current rating must be greater than 103.8W/Vsec. If it's 24V, then 97/24= 4.325A. May be 20% margin will do (ie 5.19A). Maybe make use of 24V,6A transformer or 24V,5A transformer.
This method would do for your calculation. Goodluck.