Inverter question about 12v draw

[DAS26miles]

At 50 watts of draw at 110v, how much will an inverter draw off house batteries?
I know 50 watts at 110v equal about .45 amp. What is the formula for converting it to 12v less the efficiency loss of the inverter?

[pigman1]

smkettner wrote:
Loss will be 10% to 15% so 50 watt load is ~55 watts out of the battery.

55w / 12.4v = about 4.5 amp battery load.

This is probably a reasonable estimate of inverter loss on house size inverters. Smaller ones generally are somewhat more efficient so 6-7% loss might be a good number if using a small dedicated inverter. It also depends on what you're using the loss comparisons for. If you're comparing inverter on to inverter off, you need to also include the phantom loads that you'll pick up when you turn the inverter on. These are what the coach builder hard wires into the 110V system and doesn't tell you what they are and where you MAY be able to go to shut them off. In many cases, these can add up to another 10% loss when that big inverter is turned on.

[dahkota]

My dish receiver (211Z or hopper) and 45" tv pulls about 9 amps. My dish receiver and 32" tv pulls about 7 amps.

[time2roll]

Loss will be 10% to 15% so 50 watt load is ~55 watts out of the battery.

55w / 12.4v = about 4.5 amp battery load.

[2oldman]

DAS26miles wrote:
Trying to calculate amp used in an evening of sat TV maybe 3 or 4 hours.

My tv watching runs about 10 amps. DTV HD receiver, computer speakers/amp, and Sony LCD TV.

[DAS26miles]

5 amps I can live with. It's for the Dish receiver. Trying to calculate amp used in an evening of sat TV maybe 3 or 4 hours. So maybe 25ahr used per night. Have 2 group 27 batteries so as long as I keep it under 100 amps til morning I'm fine.

[pianotuna]

Quick and dirty divide the wattage by 10. So 50 watts may cost you 5 amps @ 12 volts dc.

[naturist]

What it takes to run the inverter is a function of how big the inverter is, whether PSW or MSW, and varies with load. By that, I think there is a certain loss even if there is no load, plus there is an inefficiency loss proportional to the load the inverter is serving. I wouldn't doubt that there is also a difference from manufacturer to manufacturer, but I don't know what any of these are, and I don't know how to go about guesstimating the sum of them for an arbitrary and unspecified inverter.

As for the equation, it is volts x amps = volts x amps. (110 volts) x (.45 amps) = (12 volts) x (solve for amps) OR (110 volts) x (.45 amps) / (12 volts) = (solve for amps).

[2oldman]

About 10x the amps at 120v. 0.45 x 10 = 4.5a, plus whatever it takes to run the inverter.

X * 12v = 50w, solve for X. (about)