Output voltage calculation for full wave bridge rectifier
You should get a \$\frac{120 V_{RMS}}{36} = 3.3 V_{RMS}\$ sine wave, or \$V_{PEAK} = 3.3*\sqrt{2} = 4.67 V_{PK} \$ at the output of the transformer, since it is a 36:1 step down transformer. This is then full wave rectified, which means that each half of the sinewave period goes from 0 to 4.67V, instead of spanning from 0 to 4.67 during the first half of the cycle, followed by 0 to -4.67V for the second half of the sinewave cycle.  Since there are 2 diode drops through the full wave rectifier, the rectified output should be about 1.2V less than 4.67V, or about 3.4V.