Wire Gauge Required Question

lorelec wrote:
Sorry, Ken, I misread your example from the very beginning. If the resistance stays constant, then yes, power is directly proportional to the square of the voltage drop across it.

I was assuming your analogy was related to the power loss in a wire, which is seldom related to applied voltage, but rather to the resistance/size of the wire or the current through it, which causes the voltage drop across it.

power loss in a wire also follows the same square law change. again, if R is held constant, then the power drop is related to current squared or voltage drop across the wire squared. double the current in a wire and the power loss in the wire will quadruple. (I^2)R

same thing if you use P=V*I. in the above case doubling the current will double the voltage drop (E=I*R) so V and I both double and power quadruples.

Or P=V^2/R.

Thats why things link incandesent light bulbs change brightness so dramatically with a small voltage change.

Now, when we start talking about solar panels or MPPT controller, these are NOT linear devices and power analysis is much more complicated. But there's not much we can do about that to change performance anyway so it is what it is.


also when using % changes, you have to be careful on what numbers are used in the calculations. for example. If i drop voltage by 10% and want to know what effect that has on power change, you don't use 10% in the power calculation you use .90 (90%) E.g. for a 10% drop in current, the power changes to (I^2R) is .90^2=.81 or 81% of the base power, or a drop of 19%. not .10^2 =.01

Now if you increase current by 10%, power =1.1^r= 1.21. so for the same 10% change, power change is 21%. not 19%. Need to pay carefull attention increase/decrease % as well.