No, the resistor is the loop resistance of the twisted pair. The load is the 6 watt load. 6 watt constant load. 10-48 volts it draws 6 watts.
From: Bill Prince Sent: Thursday, March 09, 2017 8:00 PM To: [email protected] Subject: Re: [AFMUG] Ohms law Then the resistor is not the load (which is what you were implying). Something has to be dynamic, otherwise the values you gave don't add up. So what you're saying is that there is another (dynamic?) load in addition to the load of the resistor? If the power is, indeed, 6 watts, and the voltage is, indeed, 48 volts, then the load is 284 ohms in addition to this 100 ohm resistor. Otherwise, I need to know which values are dynamic. bp <part15sbs{at}gmail{dot}com> On 3/9/2017 5:27 PM, Chuck McCown wrote: Bill, 6 != I * 48 due to the voltage drop of the resistor.� So it is less than 48.� I specified the power of the load, but the power dissipated by the resistor is not specified.� There is more power than 6 watts consumed due to the resistor.� � But the more the resistor drops� the voltage, the more current the load consumes, and then there is more current through the resistor which causes more voltage drop etc etc.� � You can solve with numerical methods, but there are solutions that involve a complex number hinting that something blew up and did not converge.� � Looks simple...� looks simple... � From: Bill Prince Sent: Thursday, March 09, 2017 6:16 PM To: [email protected] Subject: Re: [AFMUG] Ohms law � This is what I sent originally: bp <part15sbs{at}gmail{dot}com> On 3/9/2017 5:06 PM, Bill Prince wrote: Iiiiieee!! What happened to that? bp <part15sbs{at}gmail{dot}com> On 3/9/2017 5:04 PM, Bill Prince wrote: Something is not right there unless the 48 volt power supply can be drawn down. Ohm's law is two equations. P = I*E�� and� E = I*R In words power equals current times voltage and voltage equals current times resistance. P = I*E�� ( 6 = I * 48 ) Solve for I�� ( 6/48 = I ) = .125 amps E = I*R ( 48 = .125 * R ) Solve for R� ( 48/.125 = R ) = 384 oms E = I*R� ( 48 = I * 100 ) Solve for I� ( 48/100 = I ) = .48 amps You can make it work if the 100 ohm load pulls the power supply voltage down. � bp <part15sbs{at}gmail{dot}com> On 3/9/2017 3:05 PM, Chuck McCown wrote: 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.�
