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Please help me understand the calculation in this table. The writer has a 12V batery connected to a 12V/110V inverter connected to standard house-hold AC appliances.

The calculation for the 1000W induction stovetop is:

1000W/12/0.85=98. 98*0.75h=73.5Ah (where 0.85 is 15% power conversion loss.)

Shouldn't it be 1000W/110V/0.85=10.7. 10.7*0.75h=8Ah?

JRE
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grunt
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  • Your question alone shows your misunderstanding, it should be "Watts of an AC device to amp-hours of a battery", and there you would have an answer to your confusion. – Edin Fifić Mar 27 '23 at 15:21
  • Note that AC power in North America is 120V not 110V. It hasn't been 110V since Vietnam. It matters on this kind of calc, since it's 9%! – Harper - Reinstate Monica Mar 28 '23 at 04:19

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If you were running the stove-top from AC you would use 120 V to figure the current, but they are calculating the Ah drawn from the battery so they use 12 V.

If you drew 8 A from a 12 V battery for an hour that would only be 96 Watt-hours, hardly what you'd expect for a stove.

So, you need 1000 W for the stove, to get 1000 W at 12 V the current would be $$ \frac{1000W}{12V} = 83.333A$$ Running the stove for 45 minutes it would use $$83.333A\times0.75h=62.5Ah$$ Accounting for 85% efficiency $$\frac{62.5Ah}{0.85}=75.53Ah$$ Which agrees with their calculation.

GodJihyo
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