Your not going to derive it for us? On Thu, Mar 9, 2017 at 9:50 PM, Chuck McCown <ch...@wbmfg.com> wrote:
> I should say, the current is the third root. > > *From:* Chuck McCown > *Sent:* Thursday, March 09, 2017 8:46 PM > *To:* af@afmug.com > *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:* af@afmug.com > *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:* af@afmug.com > *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:* af@afmug.com > *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:* af@afmug.com > *Subject:* Re: [AFMUG] Ohms law > � > > This is straight out of my Dc circuit analysis book from college. > > https://www.allaboutcircuits.com/textbook/direct-current/ > chpt-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:* af@afmug.com > *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:* af@afmug.com > *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.� > > > -- > > > -- > > > -- > > > -- >
