So I was correct...lol On Mar 10, 2017 8:05 AM, "Chuck McCown" <[email protected]> wrote:
> Screw that... > > http://www.calculatorsoup.com/calculators/algebra/cubicequation.php > > > *From:* Chuck Macenski > *Sent:* Thursday, March 09, 2017 9:27 PM > *To:* [email protected] > *Subject:* Re: [AFMUG] Ohms law > > Your not going to derive it for us? > > On Thu, Mar 9, 2017 at 9:50 PM, Chuck McCown <[email protected]> wrote: > >> I should say, the current is the third root. >> >> *From:* Chuck McCown >> *Sent:* Thursday, March 09, 2017 8:46 PM >> *To:* [email protected] >> *Subject:* Re: [AFMUG] Ohms law >> >> The solution is the third root of: 0=ax^3 + bx^2 + cx + d >> >> Where: >> Where A = Resistance >> B=0 >> C = - Voltage >> D = Power >> >> This is a third order polynomial or cubic function. It produces complex >> roots in areas where there is no solution. Like my example if you go over >> 12 watts on the load the equation goes complex. >> >> Like I said, it looks simple ... looks simple. Once I realized it was >> not linear it was much easier to arrive at the solution. Initially I >> thought quadratic but when I could not get rid of the current cubed term >> then the light bulb went on. >> >> https://en.wikipedia.org/wiki/Cubic_function >> >> >> *From:* Chuck McCown >> *Sent:* Thursday, March 09, 2017 8:35 PM >> *To:* [email protected] >> *Subject:* Re: [AFMUG] Ohms law >> >> Constants are the resistor, the 48 volt power supply, and the power >> consumed by the load. >> Yes, the load resistance changes. Many devices are like this. Canopy SMs >> are like this. >> >> 48^2/6 only works if there is zero voltage drop across the loop (100 ohm >> resistor). >> >> It is 48-(I^2R)/6 >> >> *From:* David Milholen >> *Sent:* Thursday, March 09, 2017 8:20 PM >> *To:* [email protected] >> *Subject:* Re: [AFMUG] Ohms law >> >> >> Ah.. Now I get it.. >> >> So, the constants are the resistor and the Power used. >> >> source E^2/P=R Load >> >> 48^2/6=384 Ohms >> >> So, if the load power remains the same throughout any source voltage >> changes then the Load resistance will change. >> >> Bill, is on the same rail I am there has to be a dynamic whether it be >> loss in the loop or loss in the load. >> >> In dc I would believe that the load resistance has to change as copper >> wire may change in extreme conditions like lots of heat or cold. >> >> >> >> On 3/9/2017 8:34 PM, Chuck McCown wrote: >> >> The voltage of the power supply is constant.� 48 volts. >> The resistor is a fixed and constant 100 ohms. >> The load is a fixed and constant 6 watts load. >> � >> But the current of the load will change depending on the voltage applied >> to the load. >> And any change in current causes a change of voltage applied to the load >> due to a change of voltage dropped across the resistor.� >> � >> The VDSL2 ethernet range extender uses 6 watts.� It can be powered from >> 10 to 48 volts.� >> � >> This is a real world application, not just an academic exercise to tease >> everyone.� I really thought I would have the solution knocked out in 3 >> minutes.� Took about an hour. >> � >> *From:* David Milholen >> *Sent:* Thursday, March 09, 2017 7:29 PM >> *To:* [email protected] >> *Subject:* Re: [AFMUG] Ohms law >> � >> >> The idea would be to remove the power source entirely and find the >> thevenin resistance. >> >> So is the resistance changing or is the voltage? >> >> � >> On 3/9/2017 7:29 PM, Chuck McCown wrote: >> >> But what is the Thevenin equivalent of a constant POWER component? >> � >> Read my reply to Bill, there is feedback.� The voltage drop of the >> resistor changes the current drawn by the load which changes the voltage >> drop of the resistor and so on.� >> � >> *From:* David Milholen >> *Sent:* Thursday, March 09, 2017 6:24 PM >> *To:* [email protected] >> *Subject:* Re: [AFMUG] Ohms law >> � >> >> This is straight out of my Dc circuit analysis book from college. >> >> https://www.allaboutcircuits.com/textbook/direct-current/chp >> t-10/thevenins-theorem/ >> >> I kinda thought my voltage was a bit off but the 100ohm is not a fixed >> value. >> >> I just used Watts/volts to get my current. W/V=I >> >> I then used the current to get the voltage drop across the loop >> >> I*R=V drop >> >> Its been a while for the thevenins therom but if i do a little study I >> think I could get it. >> >> � >> On 3/9/2017 5:28 PM, Chuck McCown wrote: >> >> But what is the formula? >> � >> *From:* Dave >> *Sent:* Thursday, March 09, 2017 4:20 PM >> *To:* [email protected] >> *Subject:* Re: [AFMUG] Ohms law >> � >> Current =.125A at load >> Voltage=35.5 at load >> >> If my current is correct then I should be on point. >> Otherwise I would use Thevenins Therom to get closer. >> >> >> On 03/09/2017 05:08 PM, Chuck McCown wrote: >> >> The questions are: >> What is the current and voltage on the load.� >> � >> *From:* Chuck McCown >> *Sent:* Thursday, March 09, 2017 4:05 PM >> *To:* [email protected] >> *Subject:* [AFMUG] Ohms law >> � >> Had a fun afternoon.� >> � >> Solve this.... and give the general formula... >> � >> 48 volt power supply >> 100 ohm wire resistance to the load. >> 6 watt load. >> � >> Took me some time.� Not trivial.� >> >> >> -- >> >> >> -- >> >> >> -- >> >> >> -- >> > >
