profdant139 wrote:
Ed, I am the OP, and I am again showing my ignorance here -- why would a 120 wat portable panel (like mine) be immune to a voltage drop? Is the drop worse for more powerful solar panels?
Voltage drop is a direct application of Ohm's Law, V=IR. The voltage across a resistance will equal the current through it multiplied by it's resistance; and given any two of them, the third can be thusly computed.
For a 12 gauge wire, for example, the resistance is (based on a quick web search) 1.588 ohms per 1000', or equivalently 1.588 milliohms per foot. A 50 foot length to the solar panel, with 100' total wire (50' for the positive and 50' for the negative), would have a resistance of roughly 0.16 ohms. The 120 watt panel nominally produces 10A at 12V, so the voltage drop would be 10A * 0.16 ohms = 1.6V worst case.
A higher power system, say 480 watts, would produce more current (40A) and--ignoring the fact that this may well exceed the ampacity of the 12 gauge wire--the voltage drop would be correspondingly greater, at 6.4V. However, if the voltage were higher, and a controller or whatever at the far end of the wire converting it to whatever is appropriate, the current would be lower, and the voltage drop also lower, and the fraction of the loss much, much less.
For what it's worth, an Anderson Powerpole connector for 12 gauge wire has a rated resistance of 0.6 milliohms per contact, or about the same as a few inches of wire. It will not contribute significantly to the voltage drop so long as it's properly installed, kept clean, and generally working properly.
