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/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:* [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.�
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