Hello, I will attempt to give you my perspective on this.
The simple answer is that it depends upon what you are trying to do and the current costs of the components. But below is a more detailed look at what the performance differences will be.
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.
When dealing with a remote system of a size suitable to power a cabin with a limited power source, it is likely cheaper to use a higher voltage system than to increase the generating capacity, or wire size to compensate for the losses. If all you want is a light or two, then a 12V system is likely cheaper especially if the power source is nearby.
