Roger Stockton quoted:
"For a cyclically alternating electric current, RMS is equal to the
value of the direct current that would produce the same power
dissipation in a resistive load."
 - <https://en.wikipedia.org/wiki/Root_mean_square>

Roger,
You used the wrong definition. The Wikipedia quote is about the power
delivered to the *load* but the discussion was about the power losses in
the *wire* feeding the load.
The situation is the same problem as with (bad) Power Factor: even
though the average power delivered to the load with bad power factor is
the same, the short and high spikes of current will heat the wires more
than in the case of a good power factor, simply because the loss in the
wire goes with the current squared, so higher peaks (and equal *average*
current) still results in higher losses in the line due to the squaring
of the current peaks in the formula for the power loss.

To use a simple and somewhat excessive example as illustration:
Case A: DC power to an (arbitrary) load: 1 Amp continuous, 1 Ohm line
resistance so 1 Volt drop, meaning 1V * 1A = 1Watt of loss in the line.

Case B: power provided with 10% duty cycle so 10A during 10% of the
period and 0 during 90% of the period. Average current still 1A and same
power delivered to the load as in case A.
However, the line load is still 1 Ohm so the 10A current causes a 10V
drop and thus 100 Watt power is lost in the line during the 10% that the
current is flowing, the average power loss in the line is therefor: 100W
* 10% = 10W
so the average power loss in the line is 10 times as high as in case A
due to the peak current being 10 times as high, even though it is only
flowing 1/10th of the time.

(Note that in case B it is claimed that the same power is delivered to
the load, so this means that the extra power loss in the line must be
delivered by a higher power draw from the source, we are after all
dealing with physics here)

Regards,
Cor.
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