I've had that question. I don't do math very well. How much less amp draw?

All I know is a few formulas, so decided to see if the Rated Input (w) from the Technical Specs page would work in my Power formula (P=EI) for comparison of amp draw for the 115v model and the 230v model.

**115v model**
P=EI

P(power in watts) = E(volts) x I (amps)

So I = P/E

I = 1142w*/115v

*

*1142w is the Rated Input (w) from tech specs page for this 115v model*

https://www.emersonquietkool.com/ttw-2/eatc12re1/ Scroll down to tech specs.

I = 1142w/115v, or 9.93 amps

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**230v model**
Using the same above formula:

I = P/E

I = 1142w*/230v

*

*1142 is also the Rated Input (w) from tech specs page for this 230v model* https://www.emersonquietkool.com/ttw-2/eatc12re2/ Scroll down to tech specs.

I = 1142/230v, or 4.96 amps

I have the 230v 12K BTU model, and the data label on the right side of the unit says: "5.2/56 Amps, so since my calculation of the 230v unit is 4.96 amps, does this comparison of amps work? The data label on the 115v model will read approximately 9.93 amps per the 115v calculation above?

In other words, 230v unit is less amp draw, because of higher voltage and same Rated Input (w) as the 115v model, and not because of same resistance as I first mentioned, or does this comparison not work at all?