The problem is a half-wave rectifier with a 1k-ohm load operates from 120V rms 60Hz through 10 to 1 step down transformer. The diode has constant voltage drop of .7 V for any current.
Find peak voltage of rectified output.

The AC voltage is stepped down to 120/10 = 12V rms

The peak rectified output voltage should be 12 - .7 = 11.3V. The answer given is 16.27V. Can someone please point me in the right direction?

Thanks.

 

as far as I know you must use the peak voltage not the RMS voltage to calcuate the peak rectified output.

find peak voltage first : 12V * square root of 2 (1.41421356) = 16. 97V

Then find peak voltage of rectified output

16.97 - .7 = 16.27V

I hope this makes sense?

Kone