liveforphysics
Banned
Sparky52, I like you dude, but resistance does NOT cause ANY voltage drop untill a circut has loading. Running AC through wires near eachother causes inductive loading (and capactive), which is the cause of the voltage drop.
An unloaded wire will NOT have voltage drop unless its getting SOME type of loading. Any other EE's here on the board are going to be haveing a good laugh at your above concept missunderstanding. Your observations are correct, but your thought of the cause is incorrect, dont feel bad, its something that a wire runner doesnt need to understand to do there job.
BTW, I write my posts with the least technical jargon I can manage. I take out a bunch of things from them, and rewrite stuff to try to make it understandable for people who arent specialized. I'm sorry if they are still too complex. If you have a better/more simple way to describe inductive and capactive loading that is created anytime a wire is near another wire for a long distance I would love to know it.
Thats right, I cant spell, I also dont know left from right, I have trouble seeing and remembering colors, I have insomnia, I deal with mental stability issues, I have spots in my vision and a lot of other problems. I would love to see how it relates to me not haveing a strong passion for physics.
BTW, my calc reactors fit my tiny budget, and they have never had a problem. I am proud of them. IMO form follows function.
I know you are a pro wire runner, but you have a missunderstanding of what is causeing the voltage drop in the wire. If you dont belive me, here is a 3rd party voltage calculator.
http://www.csgnetwork.com/voltagedropcalc.html
enter in 1000ft for wire legnth, and use .000012amp load to represent at 10Mohm resistance (standard for DMMs). What voltage drop does it give you? 0 volts (too low of a drop for the precision of the calculator and measureing tools to matter).
If you do this is real life with DC, you will measure on your own meter that it has zero voltage drop. This is because DC doesnt have any inductive losses, but it has the exact same wire resistance.
Now, connect it to AC and measure your 1000ft wire, and I belive you that it has a 4volt drop. But its not because of the wire resistance, its because of inductive/capacitive loading.
I respect your oppinion, and its clear you are a pro wire runner, however, no load means no voltage drop. Ask any fellow EE this, they will tell you the same thing. If you measure a voltage drop, they will say something like, well, its wasnt unloaded then, it had inductive or capacitive loading.
Im sure you know that when you just run wires next to eachother, and/or in a grounded metal tube or whatever that you are loading the circut, because you have created a capacitor, which is useing current to charge and discharge it every cycle.
I think you dont belive me, and that you dont trust EE's (like most electricians). So, try the same 1000' test out with DC and notice that you will have zero voltage drop. Was the wire resistance any less? Nope, its because its not loading the circut through inductive and capacitive means.
I like you reguardless, and I enjoy your contributions.
-Luke
An unloaded wire will NOT have voltage drop unless its getting SOME type of loading. Any other EE's here on the board are going to be haveing a good laugh at your above concept missunderstanding. Your observations are correct, but your thought of the cause is incorrect, dont feel bad, its something that a wire runner doesnt need to understand to do there job.
BTW, I write my posts with the least technical jargon I can manage. I take out a bunch of things from them, and rewrite stuff to try to make it understandable for people who arent specialized. I'm sorry if they are still too complex. If you have a better/more simple way to describe inductive and capactive loading that is created anytime a wire is near another wire for a long distance I would love to know it.
Thats right, I cant spell, I also dont know left from right, I have trouble seeing and remembering colors, I have insomnia, I deal with mental stability issues, I have spots in my vision and a lot of other problems. I would love to see how it relates to me not haveing a strong passion for physics.
BTW, my calc reactors fit my tiny budget, and they have never had a problem. I am proud of them. IMO form follows function.
I know you are a pro wire runner, but you have a missunderstanding of what is causeing the voltage drop in the wire. If you dont belive me, here is a 3rd party voltage calculator.
http://www.csgnetwork.com/voltagedropcalc.html
enter in 1000ft for wire legnth, and use .000012amp load to represent at 10Mohm resistance (standard for DMMs). What voltage drop does it give you? 0 volts (too low of a drop for the precision of the calculator and measureing tools to matter).
If you do this is real life with DC, you will measure on your own meter that it has zero voltage drop. This is because DC doesnt have any inductive losses, but it has the exact same wire resistance.
Now, connect it to AC and measure your 1000ft wire, and I belive you that it has a 4volt drop. But its not because of the wire resistance, its because of inductive/capacitive loading.
I respect your oppinion, and its clear you are a pro wire runner, however, no load means no voltage drop. Ask any fellow EE this, they will tell you the same thing. If you measure a voltage drop, they will say something like, well, its wasnt unloaded then, it had inductive or capacitive loading.
Im sure you know that when you just run wires next to eachother, and/or in a grounded metal tube or whatever that you are loading the circut, because you have created a capacitor, which is useing current to charge and discharge it every cycle.
I think you dont belive me, and that you dont trust EE's (like most electricians). So, try the same 1000' test out with DC and notice that you will have zero voltage drop. Was the wire resistance any less? Nope, its because its not loading the circut through inductive and capacitive means.
I like you reguardless, and I enjoy your contributions.
-Luke
Last edited:
