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I have a DC to AC power inverter with the following information on the specs sticker:

Input: 12 VDC, 50 A max

AC Output: 115 VAC, 60 Hz, 400 W max (5 min) / 320 W continuous

Given that current in amps is equal to the division of power in watts by voltage in volts, I(A) = P(W) / V(V), is the calculation below correct?

50 A = P(W) / 12 VDC

50 A * 12 VDC = 600 W

And since the information about continuous DC power is given as 320 W, is the calculation for DC to AC power conversion efficiency below correct?

320 W / 600 W = 0.53 efficiency

If the calculations above are wrong, could you please explain and correct me?

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    \$\begingroup\$ What is the make/model of the inverter? The efficiency should be in the spec. 50A is the peak power draw, not continuous, so can't be used to calculate the efficiency. \$\endgroup\$
    – LShaver
    Commented Nov 15, 2022 at 22:21
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    \$\begingroup\$ The best estimate is to assume that the 50A maximum is drawn when the inverter is providing its naximum which is 400 watts. This results in an efficiency of 400/600 or 66%. \$\endgroup\$
    – Barry
    Commented Nov 15, 2022 at 22:56
  • \$\begingroup\$ @LShaver The inverter is a Duracell DRINV400, and unfortunately I don't see any efficiency information. \$\endgroup\$
    – u20210512
    Commented Nov 16, 2022 at 6:45
  • \$\begingroup\$ Do you have one? There should be a marking on it somewhere (near things like FCC disclosures, UL listing, etc) that is a Roman numeral (usually V or VI) that tells you the efficiency class. \$\endgroup\$
    – LShaver
    Commented Nov 17, 2022 at 18:30

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The calculations are not wrong, but you may be misinterpreting the meaning of these specifications

  1. The 12VDC input is the nominal voltage, an input voltage range or tolerance should be specified.
  2. Input current will be highest at full load (400W), not continuous load, and when the input voltage is at the low end of the range. Even under these conditions, it likely won't be as high as 50A.
  3. The 50A max spec is a worst-case value, useful for sizing the power source, including fuses, wiring, switches, etc.
  4. The efficiency will vary with load.

53% efficiency (or even 67%, at 400W output) is a rather low value for full load. If efficiency is not specified, the best way to check is to actually measure input and output voltages and currents at the loads that are of interest.

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