Both roots prove. So which one will it actually be? From: [email protected] Sent: Friday, March 10, 2017 9:49 AM To: [email protected] Subject: Re: [AFMUG] the solution
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.
