Wire Gauge Required Question

BFL13 wrote:
I think of it in a different way.

When measuring for R on a wire, as taught by Salvo, you know the current and then you take the voltage between the ends using extra long probes on your voltmeter (I have speaker wire for that so I can do a 30ft wire, for instance.)

eg, you can find 0.13v ("voltage drop") with 20.6a so R = 0.0063, or you might get
0.16v with 27a so R is 0.0059 (R is the same but testing margin of error there)

The solar panel is a current maker, but the current varies with sunlight as well as the voltage, but the thing that shows on the calculator is the wire does not change--same R no matter what the current. But the voltage drop does change as the current changes- is more with higher amps.

I am not doing it right somehow for getting the change in amps at the battery for a particular voltage drop from panel-controller. You use the percentage of the voltage drop wrt panel rated watts? And that is your loss in watts at the controller? Then you have say 13.5v at the battery. Now what again? :( Just hopeless at this stuff.

remember, power is related to Voltage SQUARED or Current SQUARED. so when it comes to voltage drop or current drop, do NOT use % drop to determine what effect it has on your ouput power from an MPPT controller!!!.

you need to take the percentage change SQUARED. for example, suppose you have a 2% voltage drop. that corresponds to a 4% POWER LOSS. (0.98x.098= .96) a 5% voltage drop is a 10% POWER LOSS. a 10% voltage drop is a 19% POWER LOSS. When feeding a MPPT controller you will see this amount as a power loss on the ouput in reduced total watts or current.

So, even a 2% voltage drop, which doesn't seem like much is really a 4% loss in output POWER.

I believe it is much easier to understand by just calculating or measuring the voltage drop and calculating or measuring the panel output current and calculate the power loss in the cable as Voltage squared/R or current squared xR. That is the power decrease you will see at the output of the MPPT controller due to cable power loss. then using whatever you think is a reasonable charge voltage at that current calcuate the current change as power/voltage.

Power to the MPPT controller input = panel output power - power lost in the cable. Power at the MPPT controller output = power at the input to the MPPT controller x conversion efficiency.

In my calculations I just assume efficiency is high and doesn't change enough to make a difference with a small % change in input power. In that case ouput power drop = input power drop.

Now you are absolutely correct that at as panel output current goes down, voltage drop goes down, and power loss goes down as well. and guess what, it also follows the square law relationship so cutting the current in half will reduce the power loss not by 50% but by 75%.

And yes, your measurement method for determining resistance is very valid. However, unless the current is very high or resitance is very high, measurement resolution makes it hard to get real accurate values. It does tell you "good" or "bad" but not as much detail about "how good" or "how bad".