Wire Gauge Required Question
For some reason I just can't understand the various wire gauge tables, and I have tried to, so I am hoping somebody will take pity and do all the work! :) One table that I can follow says that for...
Forum Discussion
We have to remember when we're talking about power loss through a wire (or resistor) that it's the voltage drop *across* the wire (or resistor) that produces the power loss. If we have 1 volt at one end of the wire, and 0.98 volt at the other end, that's a 0.02 volt drop => a 2% drop in voltage, which becomes a 0.0004w loss ((0.02)^2 / 1) across a 1 ohm resistance. To produce a 0.02v drop across a 1 ohm resistance, there must be 0.02amp of current flowing through it (0.02 / 1), which means that we have 0.02w of power available. A 0.0004w loss out of 0.02w is 2% (0.0004 / 0.02). So a 2% voltage drop across the wire results in a 2% power loss in the wire.
For a constant current source, such as a solar panel operating to the left of the Vmp on the I-V curve, power loss in a wire is going to be constant so long as the current remains constant. That's why it's usually preferable to put panels in series when adding power, so that voltage and power increase but current remains the same (ideally).
For a constant current source, such as a solar panel operating to the left of the Vmp on the I-V curve, power loss in a wire is going to be constant so long as the current remains constant. That's why it's usually preferable to put panels in series when adding power, so that voltage and power increase but current remains the same (ideally).
