Electronic – Frequency of output of full-wave rectifier

The ripple formula you have is an approximation and just to demonstrate that here's another: -

The formula used here is not too disimilar from yours but it more accurately shows the time and not the frequency as being the important factor. However, the article makes an error in stating the 10msecs should be used at 50Hz. When the diode stops conducting at the top of the cycle and when it restarts is slightly less than 10msecs.

But, in the article's credit, look at the final paragraph - small print indicating where problems with the formula might lie and of course the OP's example falls into this area where all bets are off.

In truth the decay of the voltage is exponential from the top of the peak and not-linear and this will make a difference too.

Well, at a current \$\frac{1}{\sqrt2}\$ lower your \$I^2R\$ heating will be half, so the overall copper heating would be the same as with the bridge.

Win Hill et al. are simplifying things a bit here by ignoring the diode drops. In the case of a requirement to get a very low voltage where the diode drops make significant difference, and/or if the granularity of the lamination size selection leads to a transformer that has more capability than required, the centre-tapped configuration may make a lot of sense.

Also, not all the transformer losses are copper losses, so doubling copper losses does not increase the total transformer losses by 2:1, especially with cheap lams.