# Calculating W DC to W AC inverter efficiency

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?

• 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. Commented Nov 15, 2022 at 22:21
• 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%. Commented Nov 15, 2022 at 22:56
• @LShaver The inverter is a Duracell DRINV400, and unfortunately I don't see any efficiency information. Commented Nov 16, 2022 at 6:45
• 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. Commented Nov 17, 2022 at 18:30