Figuring Inverter Amperage

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To my smarter than me mathematicians in TF World.

Scenario:

1. battery bank 12VDC

2. Inverter

3. A 120AC duel outlet

Battery bank connected to inverter, which is connected to the 120 outlet.

Question:

I have a 10 amp A/C load on the 120vAC outlet.

What would be the DC amp load on the house battery bank? :confused::socool:
 
Figure somewhere around 110 amps DC. Basically, multiply by 10, then add some for inverter inefficiency.
 
Figure somewhere around 110 amps DC. Basically, multiply by 10, then add some for inverter inefficiency.

Sounds about right! Inverters use 5 to 10% more power than your pulling. As mentioned, inefficiency or loss in the conversion.
 
To my smarter than me mathematicians in TF World.

Scenario:

1. battery bank 12VDC

2. Inverter

3. A 120AC duel outlet

Battery bank connected to inverter, which is connected to the 120 outlet.

Question:

I have a 10 amp A/C load on the 120vAC outlet.

What would be the DC amp load on the house battery bank? :confused::socool:


Rough numbers and ignoring losses, your load is:
10 amps * 120 volts AC = 1200 watts

Using your batteries through your inverter for this load, you need:
1200 watts / 12 volts = 100 amps from your batteries.

Apologies to engineers and electricians. I'm only an accountant. :eek:
 
Pulling 100A is going to require a LARGE battery bank.

A 200A battery is not going to provide 100A for 30 minutes.
 
Keep in mind that most times this is not a continuous load. Take a 1,000 watt coffee maker (1000 watts / 12 volts = 83 amp hours) for example. They make the coffee (takes maybe 5 minutes) then cycle on and off to keep it warm. So over the course of an hour it is only drawing that 83 amp hours for 30 minutes. So actual draw over an hour would be 42 amp hours or so. A Victron monitor or the like will greatly help you calculate the amps used.
 
I would agree between 110 and 120 amps depending on inverter efficiency, whether pure or modified sine wave (and how the appliance handles the modified sine wave), and bank size. The larger the bank, generally, the higher the load voltage. As the bank load voltage drops, necessarily, the amp draw will increase to produce the same 120 VAC amperage. Duration of load will have a significant impact. Microwaving for 1 minute versus 10 minutes will generally show a significant increase in amperage draw by the end of 10 minutes.

Ted
 
Just order them from Greg's marine wire. Measure, specify the color you prefer, the fittings, and they come perfectly swaged with shrink wrapped terminals for about the price you'll pay for the wire. Never do my own again.
 
Inverters are most efficient at modest loads , it goes down with a tiny load .

Most inverter mfg. have a chart that shows efficiency at different loads.
 
To my smarter than me mathematicians in TF World.

Scenario:

1. battery bank 12VDC

2. Inverter

3. A 120AC duel outlet

Battery bank connected to inverter, which is connected to the 120 outlet.

Question:

I have a 10 amp A/C load on the 120vAC outlet.

What would be the DC amp load on the house battery bank? :confused::socool:

Confess, why did you ask this question?
 
Oh heck I don't know. I am replacing an old inverter and when you are retired you think of things like this.:eek:

I was curious......:rolleyes:
And thinking about these things is educational on what types of set up might work well when wanting to use an inverter. Lithium- 24 volt bank of as many AH as you can afford really starts to shine.

In the example above:
Using your batteries through your inverter for this load, you need:
1200 watts / 12 volts = 100 amps from your batteries.

Substitute the 12 for 24

Using your batteries through your inverter for this load, you need:
1200 watts / 24 volts = 50 amps from your batteries.

I have only recently been thinking about this myself. Especially since I got the Dakota Lithium batteries for my Golf Cart. They are all they are hyped to be IMO. And then going 24 volts helps keep amp loads down.
 

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