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ale0502 · 04-01-2016, 04:43 PM

If i calculate:
380V / 8,3A = 45,7ohms (resistance of coil)
so:
(120V / sqrt(2)) / 45,7ohms = 1,85A (new current with 120V)

My motor is a 2 poles, if I re-connect the series coils all in parallel, need only the middle of voltage. So the power is:

P = (2*120V) * sqrt(2) * sqrt(3/2) * 1,85A * 0,88 * 0,83
P = 562w

If run it up to 200Hz (with Paul´s controller), can give me approximately 2,2Kw.

It´s that correct?

04-01-2016, 04:43 PM	#2729 (permalink)
ale0502 Ale Madeo Join Date: Jan 2016 Location: Bariloche Patagonia Argentina Posts: 32 Thanks: 22 Thanked 7 Times in 5 Posts	If i calculate: 380V / 8,3A = 45,7ohms (resistance of coil) so: (120V / sqrt(2)) / 45,7ohms = 1,85A (new current with 120V) My motor is a 2 poles, if I re-connect the series coils all in parallel, need only the middle of voltage. So the power is: P = (2120V) sqrt(2) * sqrt(3/2) * 1,85A * 0,88 * 0,83 P = 562w If run it up to 200Hz (with Paul´s controller), can give me approximately 2,2Kw. It´s that correct? __________________ Los que dicen que es imposible, no deberían molestar a los que lo están intentando...