OK, so since it is proven, again here it is: 0= – RI^2 + VI – P
With the coefficients being: A= –R (wire loop circuit resistance) B= V (power supply voltage) C= –P (power of the constant power load) Enter those in a quadratic solver and if it does not give complex roots it will work. The current will be the lower root. Now if someone smarter than me will please figure out how to solve for the minimum voltage that you can use. I feel in my bones it is the first derivative but I cannot understand how that could be true because the power coefficient drops out. So obviously that is not the path to the solution. However when you close in on the minimum workable voltage the quadratic roots converge to the same number. Perhaps it is a limit type of problem. From: David Milholen Sent: Friday, March 10, 2017 7:31 PM To: [email protected] Subject: Re: [AFMUG] Fw: the solution Ahh yes.. AKA the LOOP :) On 3/10/2017 9:07 AM, Chuck McCown wrote: Because of the feedback.� � From: David Milholen Sent: Friday, March 10, 2017 6:02 AM To: [email protected] Subject: Re: [AFMUG] Fw: the solution � I love this stuff I feel it still keeps me sharp in my later years :) My question is why is it NOT Linear? Would it be because of the Variables in the load and wire? � On 3/9/2017 10:03 PM, Chuck McCown wrote: -----Original Message----- From: Chuck McCown Sent: Thursday, March 09, 2017 9:00 PM To: [email protected] Subject: Re: [AFMUG] Fw: the solution This was a PDF.� I can't remember if PDF come through the list or not.� -----Original Message----- From: Chuck McCown Sent: Thursday, March 09, 2017 8:59 PM To: [email protected] Subject: [AFMUG] Fw: the solution As you can see, I actually arrived at the solution early on, but then stumbled around searching for the linear solution which does not exist. -- --
