Another stupid question by your's truely

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zappy

zappy

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If you have a 16amp appliance and 12awg wire 100' to and 100' back.The ohm's is 0.2 each way = 0.4 so IxR =6.4 volts on the wire?How can this appliance run on 6.4 volts?Or is this the volts after it comes out of the appliance?Why is it when you increase the ohm's it increases the volts shouldn't it decrease it?Or is it because volts is pressure so as you increase the ohm's it increases the pressure?:confused:
 
think of the wire as a big resistor. if you can stick one lead of your meter at one end and the other lead at the other voltage drop across the wire would be 6.4 volts. when you put 120 volts in one end of the wire you will loose 6.4 when it comes out the other end when that load kicks on. thats voltage drop

the formula is CM= 2KID/VD

CM = circular mils

2 = constant multiplier
K= 12.9 for copper conductors
I= amps
D= distance one way
VD = voltage drop. to get voltage drop multiply the voltage at the panel by a percent you would like to see. the code reccomends 5% if im not mistaken

i didnt do the math to your problem but thats the forumla for voltage drop
 
Zappy,
Given 120 Volts Source, #12 cu, 200'.
The 6 Volts is the Voltage Dropped Under Load.
You have 114 Volts left to run the appliance.

A single equation only describes a portion of the system, like Voltage Drop.
Often, one must "define the target" of an equation,
then select the appropriate equationh,
then fill in the blanks.

Multiple equations are used to describe different perspectives on the same problem.

Just keep reading electrical theory in these forums.
It gives a sense of accomplishment.
It is "Adult" fun!
It is Electrician's fun!
 
glene77is said:
The 6 Volts is the Voltage Dropped Under Load.
Click to expand...
Yep, that's it. I had the hardest time understanding why higher voltages make for lower losses in transmission lines, until I woke up to this same simple fact. The voltage drop you are talking about, and that I had a hard time seeing, is what happens in the wires between the source and the load.
 
charlie b said:
Yep, that's it. I had the hardest time understanding why higher voltages make for lower losses in transmission lines, until I woke up to this same simple fact. The voltage drop you are talking about, and that I had a hard time seeing, is what happens in the wires between the source and the load.
Click to expand...

Do I detect a hint of sarcasm? :grin:
 
your appliance is not running on 6.4 volts. It runs on 120 - voltage drop. If you did voltage drop calcution right, the appliance runs on 120-6.4 volt. Anyhow, check your voltage drop calculation again. If you increase the ohm (R), Volt will increase. So, you are correct in this concept. But one thing that you forget was that the Volt is actually Voltage drop across the ohm (R). Therefore, the appliance (or load) will see Volt at source minus voltage drop across wire. I hope that answer your question.
 
https://www.rerc.org/cart/index.php..._id=30&zenid=3b8b5f06c4c5723af3ecd7089ce0c6e9

I would recommend this source


I need to order the newer 14th edition, however, I do have the 13th edition, the book has excellent voltage drop calculation tables, which come in quite handy for 1400' circuits.

For example, in your case,

load in amp awg length

15a 8 100' for a 2% drop
15a 8 100' for a 3% drop
15a 10 100' for a 4% drop

unfortunately the book does not show a 5% table, thus using the previosly posted voltage drop calculation would be required.
 
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zappy said:
If you have a 16amp appliance and 12awg wire 100' to and 100' back.The ohm's is 0.2 each way = 0.4 so IxR =6.4 volts on the wire?How can this appliance run on 6.4 volts?Or is this the volts after it comes out of the appliance?Why is it when you increase the ohm's it increases the volts shouldn't it decrease it?Or is it because volts is pressure so as you increase the ohm's it increases the pressure?:confused:
Click to expand...

The 2 wires serving the equipment are in series with the load. So think of it droping 3.2 volts on the way out to the equipment. 113.6 volts across the load. And 3.2 volts on the way back to the source. The drop is only created by the high current draw. If the load were smaller the drop would be less.
 
090317-0923 EST

In Brian Johns post there is an assumption of a source voltage that I will guess is 120 V. You can not relate current and percentage without also defining voltage.

electricalperson:

Your constant for copper is defined at what temperature?

Working backwards from your equation
V/I = 2*D*K/CM
removing the factor of 2
V/I = D*K/CM
produces
R = 1000*12.9/10380 ohms/1000 ft for #10 solid wire
R = 1.243 ohms/1000 ft

My table from "Reference Data for Radio Engineers", Fourth Edition, International Telephone and Telegraph Corp., lists annealed copper AGW solid wire at 20 deg C as
0.9989 ohms/1000 ft.

Do you know if your K factor is defined for an elevated wire temperature? Also I believe stranded wire for a given nominal size has a slightly smaller cross sectional area. So might this K value also include a factor for this difference?

.
 
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Yea, Gar, sry, I should have entered the voltage, the publication has 30 different tables with differing voltages, copper vs alum. ,and amps . I was/am merely suggesting a publication.
 
zappy said:
If you have a 16amp appliance and 12awg wire 100' to and 100' back.The ohm's is 0.2 each way = 0.4 so IxR =6.4 volts on the wire?How can this appliance run on 6.4 volts?Or is this the volts after it comes out of the appliance?Why is it when you increase the ohm's it increases the volts shouldn't it decrease it?Or is it because volts is pressure so as you increase the ohm's it increases the pressure?:confused:
Click to expand...

One thing you might want to consider is that the appliance might not draw as much current given the voltage drop through the wiring. For example, considering a resistive load (like a heater element) of 16A @ 120V, it would be 7.5 ohm (120V/16A). Now, add .4 ohms from the wiring for a total of 7.9 ohm, and the current at 120V = 15.2A (120V/7.9ohm). So your "16A @ 120V" (1920W) appliance now is instead running "15.2A @ 114V" (1732W).

Oh, and the AWG table I refer to says .1588 ohm for 100' of #12 (I used .2 ohm in the calculations above) :)
 
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