This all started with the VDSL2 ethernet line extenders. They are a constant load of 6 watts. I want to power them over the pair of wires that the VDSL signal is going over. So the question is, if I put 48 volts on the pair at the house through a VDSL filter, can I get enough power out at the far end to power that VDSL unit. 100 ohm loop. About 2000 feet of 24 gauge. The answer is yes.
Current is 0.12953 A Voltage at the load is 46.322 V You can brute force it with trial and error, that is how I got my first estimate, but I wanted to know exactly. There is a formula. Trying to visualize how I would use Thevenin. The battery is a short, right? So from the load’s perspective you are seeing 100 ohms. Not sure what the Thevenin equivalent of a constant power load is. From: Chuck McCown Sent: Thursday, March 09, 2017 4:28 PM To: [email protected] Subject: Re: [AFMUG] Ohms law 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.� --
