This would be true if you didn't rewire the motor from a parallel
connection to a series connection when going from 110 to 220. Current,
therefore I^2R losses, should be the same. Most small motors, however
don't change the connection to the start winding so it DOES carry twice
the current during the brief period it's active.

rhg

Richard Cline wrote:

In article s.com,
(Robert Bonomi) wrote:

Keep in mind that the light bulb example is totally different than the
saw. In the light bulb, the watts go into heating the filament. A 60
watt bulb is just that regardless of the voltage/current relationship to
create the power.

In the case of the saw, most of the energy goes into cutting wood. Only
a small amount goes into heating the motor. The motor heating goes as
I^2*R. As the lower voltage requires a higher current, there is more
heating in the lower voltage situation.

Dick


In article ,
Doug Miller wrote:

In article , "BadgerDog"
wrote:

Why would the saw run cooler if there is more voltage to the saw?

Because the current is less, and it's current that generates heat, not
voltage.

Oh, really?

Which generates more heat:
1) a 60-watt lightbulb (designed for 120V) running at 120v
2) a 60-watt lightbulb (designed for 240V) running at 240v

Note that #1 is drawing 1/2 amp, and #2 is drawing only 1/4 amp.


"Watts is watts", applies -- It doesn't matter how they're produced.