Okay, so the circular logic is real and you would just keep running the VD calculation with a smaller and smaller increase in line current as you approach infinity?

I'm not crazy. Whew. So, this is what you meant by "differential equation and a headache?"

Ha ha – Jerramundi you are not crazy

And Winnie is spot on – it’s all math yo! And he is also correct when he indicates it’s not worth trying to be perfect here (

*Perfect is the enemy of Good*) so knowing when to say “

*Good Enough*” is important!

I think you see correctly that your constant power load will draw more current to compensate for the voltage drop, then that slightly higher new current means more voltage drop, which means more current drawn which means more voltage drop… and on and on.

You will find, that for your application, the “limit” that Winnie is describing is approached quite rapidly. We are not doing rocket science and I’d argue anything much beyond that first digit to the right of the decimal point (tenths) is not to lose any sleep over.

If you iterate, that is keep calculating, your new voltage drop calculations you will see that after the 2nd or 3rd iteration you settle quickly on ~ 13.26A

For fun…

Load is 1440W

1440 W load draws

__12A__ at 120V

At 271 feet and 12A, 120V drops to 109.7V............................Voltage drop calc #1

Load now draws

__13.13A__ at 109.7V

At 271 feet and 13.13A, 120V drops to 108.7V.......................Voltage drop calc #2

Load now draws

__13.25A__ at 108.7V

At 271 feet and 13.25A, 120V drops to 108.6V........................Voltage drop calc #3

Load now draws

__13.26A__ at 108.6V

You could keep recalculating this, over and over, and you’d get a new answer every time. But after the 3rd calculation, it’s silly to keep doing because the answer is changing only slightly. So iterate a few times to watch how fast you approach a limit and then know when to say,

*That’s good enough*!