Hey, I only noticed it so quickly because I went down the exact same path you did. I spent a more than an hour before going to bed frustrated (with the math). Guess you should never assume that every problem has a solution :)
On Fri, Mar 10, 2017 at 10:49 AM, <[email protected]> wrote: > I guess Chuck Macenski was the first to point out my error. > > *From:* [email protected] > *Sent:* Friday, March 10, 2017 9:47 AM > *To:* [email protected] > *Subject:* Re: [AFMUG] the solution > > Ok due to brain fart yesterday, everything was wrong. Voltage drop does *not > *equal I squared R, thanks Mark. Rusty brain cells. > > It is a quadratic. > > -RI^2 + V^I – P =0 > > Which my son William pointed out to me yesterday. I really thought I was > smarter than him... he gets a point for this. > > My 48 volt example has complex roots. So it will not work. > Just bumping the voltage up to 50 volts allows it to work. > > It solves for two roots, one at .3 and one at .2 > > The second root, .2 amps proves out. .2 amps times 100 ohms = 20 volts. > So 50 volts power supply – 20 volt resistor drop = 30 volts across the > load. > 6 watts / 30 volts = 200 Ma. That same as the second root. > So the solution “appears” to be working. > > Lets try 100 volts. > > The quadratic solver says the second root is .064 (64 mA) > > .064*100=6.4 volts > 93.6 volts on the load. > 6/93.6= 64mA Eureka! > > I was hopeful the first derivative would be the current at the minimum > workable voltage. But that did not work out on my first attempt. At the > minimum workable voltage, which is something like 48.9 volts the two roots > almost converge. Someone with a bit more math horsepower will be needed to > solve that for me. That would be a very useful number. You have a power > and a loop resistance, what is the minimum voltage needed to make it work. > > *From:* Chuck McCown > *Sent:* Friday, March 10, 2017 8:18 AM > *To:* [email protected] > *Subject:* Re: [AFMUG] the solution > > Gimme another hour... > > *From:* Mark Radabaugh > *Sent:* Friday, March 10, 2017 8:07 AM > *To:* [email protected] > *Subject:* Re: [AFMUG] the solution > > what what? > > P = i^2R > > V=IR > P=VI > > ignoring power factor on linear DC > > Mark > > > On Mar 10, 2017, at 12:07 AM, Chuck Macenski <[email protected]> wrote: > > Ok, doing software apparently has erased some important stuff from my > brain. Hard to know what else I lost. Having said that, why is Vr = I**2 * > R? Wouldn't Vr = I * R? > > On Thu, Mar 9, 2017 at 9:59 PM, Chuck McCown <[email protected]> wrote: > >> >> 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. > > > > >
