Have I calculated the peak rectified V and diode I correctly?
Assuming you mean this:







simulate this circuit &ndash; Schematic created using CircuitLab

Then you would first calculate the AC output of the transformer.

Since you have a 20 to 1 turns ratio, that'd be the input voltage divided by 20.  Use the RMS value of the input voltage.  In your example, that's 240VAC. Dividing by the turns ratio gives you the transformer output voltage as RMS.

Once you have the RMS voltage on the low voltage side of the transformer, you can figure out the peak voltage from the rectifier:


AC peak to peak: \$V_{pp} = V_{RMS} \times 2\sqrt{2}\$
Rectified peak: \$V_R = V_{pp}/2 - 2V_{f}\$
Peak current: \$I_P = V_R / R\$


Where \$V_f\$ is the forward voltage of one diode in your bridge rectifier, and \$I_P\$ is the peak current through \$R1\$.

Since you say "assume ideal diodes," \$V_f\$ would be zero.



Pretty sure that's right.  If not, someone please let me know and I'll correct it and Nexus_Valentine and I will both have learned something.