But I do know that it has a lot to do with your final intended size of your alternative energy setup.

Use 12v dc for really small systems, RV's and cabins.

Use 24 volts dc for larger of the above and small homes.

Use 48 volts dc for larger homes, shops etc.

Use 120 volts dc for extremely large homes and power needs.

Where the cut off points are on one voltage to the next I'm not sure, and maybe a Pro can tell us. But the voltage you use has every thing to do with your final anticipated load usage 20 years down the road.

It is a huge mistake to start with a low voltage system when you will eventually need a higher voltage system... It will cost you thousands of dollars for making that mistake. For instance your 12v copper wires may become as large as 0000 in size.. This is the size of garden hose and will cost a fortune. In such a case the planned voltage should have been much higher in the first place.

Also don't make the mistake of comparing inverter prices with the cheepie mass marketed inverters versus the extra heavy duty inverters that are designed to work 24 hrs a day year after year. There is a large difference between an inverter that's designed to just work one or two weekends a year out camping versus one made to run your house loads day in and day out.

In A.E. systems there are many loads that are more efficient being driven directly rather than through an inverter. So you will need dc wires to those loads.

Also redundancy is a factor. If everything on your property including your night light is dependent on one or two large inverters and those inverters die from a lighting strike, then your whole system is down... nothing will work. But if a lot of your items are able to run directly off of DC then they will keep operating even when no inverter is operational.

Even inverters can only handle a certain size of DC wire running to them. If you got a huge inverter, chances are it's got to be a higher voltage inverter, because even if it sets right next to your batteries, you'll probably not be able to find large enough wire to hook it up. And your jumpers from battery post to post would be enormous.

I submit the below example for your consideration.  Hopefully my math is correct.

First, some physics definitions:

Watts = Volts * Amps

Ohm's law, V=IR  Where V=voltage, or in the case of a conductor  this can be looked at as voltage drop.  If the Voltage is in Volts, the Current is in Amps, and the resistance is in Ohms, then you can directly plug into this equation.

Power equation, P=IV where P is the power, I is the current, and V is the voltage.  If power is in Watts, Current is in Amps, and Voltage is in Volts, you can directly plug into this equation.

Example 1  (these have been simplified for use as an example)
If one wanted to move 200 watts of power from 4 12 Volt solar panels to an inverter 100 feet away, and one had a 12 guage cable, how much power would be lost in the cables as heat for a 12V system, and for a 48V system?

For the 12 Volt system, if the panels put out 200 watts arranged in parallel at 12 volts, then from ohms law the current would be 200 = 12 * Amps, or Amps = 16.66
Next we will need to know the resistance of 200 feet of 12 guage wire, remember that it is 100 feet there and 100 feet back round trip for the electrons. We can google wire resistance chart and find a convenient chart that shows that 12 guage copper wire has a resistance of 1.620 Ohms per thousand feet.  Which means that for 200 feet, the resistance would be .324 Ohms.  plugging this back into Ohms law, this means that the voltage drop in the wire is 16.66 Amps*.324 Ohms = 5.39 volts.  The power consumed in the wire would be from the power equation equal to 16.66 Amps * 5.39 volts, or 89.79 Watts.  This would be the power that winds up as heat in the conductor.  That would be nearly half of the power generated.  

For the 48 Volt system, If the panels put out 200 watts at 48V when connected in series, then from ohms law the current would be 200 watts / 48 volts = 4.17 Amps.  The resistance of the 12 guage wire still being .324 Ohms, yeilds a voltage drop calculated from Ohm's Law of 4.17 Amps * .324 Ohms which equals 1.35 Volts.  Plugging this back into the power equation results in a power of 4.17 Amps * 1.35 Volts, or 5.62 Watts.  

In a real system, there would be additional variables as the panels would put out varying voltages depending on load, the resistance of the conductor would vary with temperature, plus many more minor variables.  It should, however, be noted that increasing the voltage by 4 times resulted in reducing the losses in the conductor by 16 times, or the square of the voltage multiplier.

This difference would become more valuable to the system designer the greater the losses in the conductor were.  Therefore, the more power you are moving around, and the further that power has to be moved, the more practical and economical a higher voltage system becomes.

In the example above, one could compensate for the voltage difference to obtain equivalent losses by using a larger conductor.  To do that one would have to replace the 12 guage wire with a larger size such that the power equation yielded a result of 5.62 watts, hence 5.62 Watts = V * 16.66 Amps.  Solving for V, the voltage drop, we obtain .337 Volts.  Plugging this into Ohm's Law and solving for th needed resistance yeilds .0202 Ohm's, or a conductor with less than .101 Ohms resistance per thousand feet.  This would require size 0 wire, which would represent a significant parts and installation cost difference when compared to 12 guage wire.  

Before this thread, I would have been tempted to answer this question, thinking that my experience with RV inverter systems has taught me all I need to know about alternative energy use of inverters.  NOT!

Interesting new look at an old topic...inverter voltage/battery bank topology is effected by your DC sources!  In an RV you rarely find yourself worrying about how many feet of cable there is running out to your wind mill.  8^)

Would using an MPPT charge controller adequately solve this part of the equation though?  I'm thinking maybe 48+ volts from  your remote wind gennie and or remote solar panels coming back to the utility shed where your MPPT charger converts it to 24V at higher current for your battery bank.

Ignoring the input DC for the moment, wouldn't a two inverter system (two 3KW 120VAC "legs" set up to also provide 240VAC) with 24VDC for each one be quite practical for a typical house?  As long as you keep the batteries close to the inverter?  In my experience with RV systems, anything below 200 amps on the DC storage side seems to wind up being pretty practical.

Taking the MPPT idea for the remote DC sources a step further...most of the big honkin inverters are inverter/chargers.  Maybe an off-the-grid system could still make use of those chargers by placing a small 1000watt inverter very near the mill and send 120Vac back to the shed?  Feed that to the AC input of the inverters.  Just a thought.  Of course some of you guys have a lot more than 1000 watts of gennie, so maybe it's not a great idea.