Why does the output voltage drop by 2VD? Whats troubling me is the ground placed after the load resistor. If we were to remove that ground and measure Vout as shown in fig 4.25a, i understand that there would be a 2VD drop. Like this though, it seems like there should only be a drop of 1VD, no?
you forgot about the the 1VD drop from the - node of the secondary winding to the GND node so you get a net of 2DV drop. Basically, the - node of vs is at 1VD below GND.
If you trace out the circuit loop, it goes through 2 diodes.