Why is There a Factor of 2 in the PIV Rating for a Full-Wave Bridge Rectifier?

[SuperCat]

Homework Statement

Part C of the problem below. I have the solution but I don't follow the logic.

Homework Equations

Turns ratio: Vp/Vs = Np / Ns
KVL
I do not think I need any elaborate formulas to solve this.

The Attempt at a Solution

Solution Provided by instructor:

I am confused why there is 2 in front of Vs.
When I apply KVL to a loop it looks something like:
Reverse Biased Diode = Forward Biased Diode + Vo

My KVL matches up with page 213 of Microelectronics Circuits Sedra 7e:

The solution provided to me seems like it would match up with a center-tapped full-wave rectifier, but I have asked a handful of classmates who all seem to agree with the provided solution so I was hoping to either receive confirmation that the solution provided to me is wrong or an elaboration.

Many thanks!

 

[Tom.G]

You are right.
You might pose this question to your instructor:
Since one end of the secondary is always connected to ground thru a conducting diode, and the only voltage source is the transformer secondary, where does the factor of 2 come from for the PIV rating?

P.S. The PIV is always the peak line-to-line voltage. The factor of 2 applies to full wave center tapped only in that the output voltage is derived from the voltage of one leg of the transformer, i.e. center tap to one end.

P.P.S. In the real world of making stuff that continues to work, a safety factor of 2 is used in almost all component ratings. I doubt that he has this in mind for this course on theory.

A rather old reference:
Reference Data for Engineers: Radio, Electronics, Computer, and Communications. Seventh Edition, 1985. Pg. 14-12. Howard W. Sams & Co. Indianapolis , Indiana, ISBN 0-672-21563-2