Hello everyone, this isn't exactly an arduino question (although I would be using arduino in the project most likely to take sensor readings and such! hehe), but I find people here to be pretty knowledgeable about electronics in general usually so wanted to ask this.
I'm theorizing about some experiments with solar power. And I'm thinking about the amperage drain on the other side of a 110 volt inverter. Do I just carry my watts over to the voltage source on the other side to calculate the amperage drain?
Example, let's say I've got a 2 amp drain on 110 volts (~220 Watts). If I have a 12v DC to 110v Inverter, does that mean my amperage drain of the 12v DC would be 220 / 12 = ~18.3 amps?
If my math is right, then surely there are some efficiency factors in the inverter that I need to take into account as well?
Side question if anyone has experimented with Solar energy before. If I had panels that were say 135 watts / 18 volts / 7.5 amps. Could I use a panel like that to charge a 12 volt dc battery? A lot of stuff I read always talks about charging a 12 volt battery with a 12 volt panel. But the panels I was thinking about using would be some more DIY monstrosities. Looking at using some of the ebay solar cells available, thinking 36 - 6x6 cells that are 3.9watts, .5 volts each. Was only considering that sizes because there are some aluminum frames and such available to readily fit that size panel. If I made a custom frame I could of course just use 24 cells instead of 36 in order to make it 12 volts. But I figured since the voltage is at peak efficieny, an 18 volt panel might be ok anyway. Obviously though, if I my amperage drain is correct above, one of these panels would not be enough to match the amperage drain 
Thanks!
If my math is right, then surely there are some efficiency factors in the inverter that I need to take into account as well?
Yes it depends on the inverter. If you do not know then you have to measure it. The efficiency will normally be between 70 to 90 %
Grumpy_Mike:
If my math is right, then surely there are some efficiency factors in the inverter that I need to take into account as well?
Yes it depends on the inverter. If you do not know then you have to measure it. The efficiency will normally be between 70 to 90 %
I'm assuming this means I am calculating it correctly at least?
What would I use to measure the efficiency? I don't think my multi-meter could handle amperages past 10 amps, I'd have to double check it but pretty sure that is the size of the high amp side fuse.
What I'm thinking about using for my inverter is an APC UPS unit. It's convenient in the fact that it should be fairly easy to rig it up to auto switch to grid power if the battery pack runs down.
I'm assuming this means I am calculating it correctly at least?
Yes the principle is right, you take the wattage and divide it up into the required current and voltage.
So if you have a supply with say 10 Watts and an 80% efficient converter you then have 8 Watts to play with.
So you could get that by 2 Amps at 4V or 4 Amps at 2V, or anything that multiplies up to 8.
What would I use to measure the efficiency?
You need to measure the input power and the output power. That means measuring the voltage and current at each side. You will need a load on the output to cause a current flow. You should do this at close to the actual voltage you require because the efficiency of an inverter changes with changing conditions.
I don't think my multi-meter could handle amperages past 10 amps
Then you need a better meter or a current probe for the meter you have.