14ga or 12ga...Which is more effiecient and why?

[LarryFine]

Hmmm, not sure I buy into that one.

A heater will run as long as it takes to satisfy the thermostat. A 1kw heater running at 900w, for instance, will have to run longer to provide the same heat gain.

If the voltage drop is causing heat inside the heated space, the results should be equity. But, if they're outside the heated space, the heat in the wires will be waste.

 

[ozark01]

I think I would put me a temporary receptacle at the panel and run a 12 gage drop cord and a 14 gage drop cord and see if you have the same problem. My guess is you have other problems other than wire size and a 14 gage drop cord run from the temp receptacle would eliminate the wire size as a problem.

 

[mivey]

if they're outside the heated space, the heat in the wires will be waste.

I'll buy that.

 

[QES]

A load in the mA, will the #14 be better?

 

[LarryFine]

A load in the mA, will the #14 be better?

Yes, if cost matters.

 

[Chamuit]

I am not smart enough to narow the questions down.

In my home office right now I have 10A of load on a 14ga wire. When I hit the print button on the computer, the lights dim, and the computer battery picks up, and the printer resets. The printer is 6 amps. I am 30 feet from the electrical panel. This is just my example of a real world situation. Not all people have this kind of load in a room. What I am trying to pry out of everyones brain is does/will/could this kind of situation, lead to premature failure of electrical equipment? I could list lots of examples of light load and heavy load situations on branch circuits. That is not what I want. I want to know if an average branch circuit is more efficient in 14 or 12. I guess it depends on your average....Average TV, Lights and a Vacume will crate this situation.

Thank you

From my perspective, you need to read what you just wrote.

Residential - using #14 - 15A breaker. Your known load is 10 amps (probably with the printer in stand-by.) You add in the 6 amps. Your load is now 16A, on a 15 amp breaker. If your office is like mine, all of the office equipment is on 24/7 which makes it a . . . continuous . . . load. So, you add 25% of the 10 amps and you are at 12.5A plus the 6A making 20. Of course, we don't know if you have any "phantom" or unknown loads either.

Our office is 9 x 12 -ish and we have 4 circuits feeding everything.

As I'm was writing this the thought comes to me that the load goes up, the slow trip of the breaker + the voltage drop allows the ups to engage and relieve the load of the computer off of the circuit creating your scenario. . . . maybe.

 

[broadgage]

A load in the mA, will the #14 be better?

No, the #12 would still give less voltage drop and therefore less waste of energy, the saving would however be minute and of no practical consequence whatsoever, unless the circuit was many miles in length.

 

[electricmanscott]

How would you sell? Wrap that up and put a bow on it....
What would you tell your grandmother if she was considering using 14 instead of 12?

If this was a math questions there would be 3 pages on how to calculate it 10 differnt ways.....

I'd say don't waste your money on some electricians misguided opinion grandma.

 

[iwire]

14ga or 12ga...Which is more effiecient and why?

I specifically want to know how a 12ga wire is more or less efficient than 14ga, and how that effects the electrical equipment utilizing that circuit?

I am looking for the dummy version of this answer. I am too technical to explain this to my customers so they can understand it. It needs to be factual with simple examples please.

Thank you for your help.

To me this is exactly the same as asking which color in the rainbow is best and why.

 

[iMuse97]

has anyone mentioned the fact that it is not efficient to waste CU at today's prices? If the load you are serving is small or intermittent, and can be served by #14AWG, it would make no sense to use #12. As to the original post:

I just use the plumbing analogy of pipe size(AWG), water flow(Current), and pressure(Voltage), and hope the customer understands, and trusts that I understand what is actually needed.

 

[tryinghard]

A ?-ton truck is used based on efficiency/economics/safety/sufficiency for a particular load range just as a 1-ton truck would be. Bigger is not prudent within this criteria nor is too small, correctly sized is.
Our charge is to understand the application and apply the size suitable resulting in efficient/economic/safe/sufficient circuitry.

 

[gar]

090919-1151 EST

Palmbay:

Is your question really about efficiency or something else?

First, consider efficiency in the sense of the ratio of useful power to input power. Also assume a constant resistance load of 10 ohms and a constant source voltage of 120 V. Four calculations, 50 ft and 100 ft for the
distances from the source voltage, and #14 (2.525 ohms/1000 ft) and #12 (1.588 ohms/1000 ft).

Disregard any slight error in my least significant digit. Also check my math for any mistakes.

50 ft distance and #14.
Copper resistance = 0.2525 ohms. Current = 120/10.2525 = 11.7045 . Power dissipated in the #14 = 34.59 W or 0.69 W per foot. Power dissipated in the load 10.2525*10.2525*10 = 1369.94 W. Input power is 120*11.7045 = 1404.54 W. Efficiency = 1369.94/1404.54 = 0.9754 or 97.54%.

100 ft distance and #14.
Copper resistance = 0.5050 ohms. Current = 120/10.5050 = 11.4231 . Power dissipated in the #14 = 65.896 W or 0.659 W per foot. Power dissipated in the load 11.4231*11.4231*10 = 1304.88 W. Input power is 120*11.4231 = 1370.78 W. Efficiency = 1304.88/1370.78 = 0.9519 or 95.19%.

So we went from a loss of 2.46% to 4.81%. Approximately doubling the losses, and reduced the output power by 2.4% .

If you had a motor load that was constant power rather than constant resistance, then loss percentage would be greater for the longer run.

50 ft #12.
Copper resistance = 0.1588 ohms. Current = 120/10.1588 = 11.8124 . Power dissipated in the #12 = 22.16 W or 0.44 W per foot. Power dissipated in the load 10.2525*10.2525*10 = 1395.33 W. Input power is 120*11.8124 = 1417.49 W. Efficiency = 1395.33/1417.49 = 0.9844 or 98.44%.

100 ft #12.
Copper resistance = 0.3176 ohms. Current = 120/10.2525 = 11.6306 . Power dissipated in the #12 = 42.96 W or 0.43 W per foot. Power dissipated in the load 10.2525*10.2525*10 = 1352.71 W. Input power is 120*11.7045 = 1395.67 W. Efficiency = 1352.71/1395.67 = 0.9692 or 96.92%.

Loss changed from of 1.66% to 3.08%. Approximately doubling the losses, and reduced the output power by 1.54% .


Consider a continuous load, then the losses in the #14 wire at 100 ft per year are 0.066*8760 = 578 KWH or about 578*0.13 = $75/year.

For #12 wire I get about 377 KWH or $49/year.

This kind of load might not be an unreasonable average load for a typical home. So is saving $25 per year on power loss worth the extra cost of the copper wire and installation. This is not a complete story, and 100 ft is probably long for an average distance for an average home. The 10 ohm continuous load is probably a good average load for a lived in home. That is an average consumption of 1.44 KWH.


Maybe more important is a question of voltage drop. At 100 ft, #12, and 10 ohm load the drop is about 0.3176/10.3176 = 3.1 % or 3.69 V. Change to #14 and the result is about 5.8 V. Both of these will cause noticeable flicker on an incandescent lamp.

Consider your computer problem. You need to make some voltage measurements. Also do some controlled voltage change measurements with the on-off of a 1500 W heater. You have a low source voltage, maybe 110 or lower (mine is about 122 to 125), and/or high resistance connections in you path from the pole transformer to your computer.

I have a UPS that trips into backup at about 100 V, then produces about 111 V, returns to non-backup at about 107 V. In other words the hysteresis is about 7 V. From memory I had one that went into backup at about 105 V or somewhat higher. That one could be a real nuisance in an area with a nominal 110 supply.

Today my voltage is about 123.8 V at my work bench and adding the 1500 W load it drops to 119.9 or a 3.9 V drop for about 12 A. This same load produces about 0.8 V change at my main panel, and most of this is the source impedance of the pole transformer.

At a more normal outlet in my house, than the bench, I get about a 1.4 V drop of which nearly 0.8 is the pole transformer drop. I have no circuit that has less than 2 breakers in series and the main fuses. The fuses have virtually no drop and the breakers are in the 0.1 V range.

You need to look at where the voltage drops are in your circuit that supplies your computer and printer.

I have no simple answer of how to explain whether to use #12 or #14 other than to say --- #12 for any receptacle and use 20 A breakers. Use #14 and 15 A for light circuits.

I am not in the business of wiring homes or businesses. So my criteria is quite different than for a typical home. For my own home the rough criteria are a main panel that supplies the 240 stuff, subpanels, a 120 circuit to outlets near each subpanel, several nearby 120 circuits including a separate 120 for each furnace. Virtually all lights and other outlets come from the subpanels. All 120 V receptacles are #12, and all lights are #14. All lights are from GE low voltage relays and wired with #14. Each bank of 3 relays is from a 15 A breaker. Every major room has two or more circuits for receptacles.

I believe your biggest problem in general is voltage drop. You could probably use #14 in most cases without great problems. Don't daisy chain thru receptacles and don't use any back stab (press in connections).

If you need to supply a 1.5 HP DeWalt radial arm saw use 240, and you could probably still use #14 over 25 to 50 ft. The same saw at 120 V and 100 ft of #12 is real marginal. These are voltage drop problems because of the high inrush current with a large diameter blade and its inertia. If I remember correctly, at the end of about 100 ft of #12 and a 120 circuit the inrush is about 70 to 80 A and lasts for several seconds.

For a kitchen it is probably better to have more #14 circuits than fewer #12. Obviously code determines what you do in this respect. Most kitchen devices would be 1500 W or less. The kitchen is probably a current limit problem rather than voltage drop. Thus, one circuit per device makes sense. You design for what the application may be and what code requires. You might very well design with more circuits than what code requires. My house was built 43 years ago and I have a separate 20 A circuit for the dishwasher, disposal, and four more circuits in the counter area. Lights are obviously separate based on my above discussion.

.

 

[Smart $]

From my perspective, you need to read what you just wrote.

Residential - using #14 - 15A breaker. Your known load is 10 amps (probably with the printer in stand-by.) You add in the 6 amps. Your load is now 16A, on a 15 amp breaker. If your office is like mine, all of the office equipment is on 24/7 which makes it a . . . continuous . . . load. So, you add 25% of the 10 amps and you are at 12.5A plus the 6A making 20. Of course, we don't know if you have any "phantom" or unknown loads either.

Our office is 9 x 12 -ish and we have 4 circuits feeding everything.

As I'm was writing this the thought comes to me that the load goes up, the slow trip of the breaker + the voltage drop allows the ups to engage and relieve the load of the computer off of the circuit creating your scenario. . . . maybe.

I agree it is likely an overload condition or an abnormally-high resistive element in the branch circuit itself, such as a less-than-solid termination somewhere. For a properly installed branch circuit, the voltage drop for the event as noted would be the effect rather than the cause. Additionally, common t/m breakers may take minutes to trip even at double the rated current. Spurts of 16 amps would likely never trip the breaker unless it was faulty. With the UPS kicking in, the breaker is even less likely to trip.

Anyway, the main reason why I even decided to reply to your post in particular is highlighted in red above. Just because office equipment is turned on 24/7, does not make them a continuous load. Practically all the equipent your are referring to do not draw their maximum current while on but not in use, and even while in use, the current draw often varies.

 

[iwire]

Just because office equipment is turned on 24/7, does not make them a continuous load.

I agree 100%.

 

[LarryFine]

I agree 100%.

Is that a continuous agreement?

 

[iwire]

Is that a continuous agreement?

Well duh. :grin:

 

[macmikeman]

I say define efficiency first. If you mean current carrying capability and the effect that has on killowatt hours registered on the meter, yes there will be a microscopic advantage to using #12. If you mean seperating the customer from his bank account for your personal use, then using 14 guage wire wherever possible and also those profit holes in the back of the devices and a crew of home depot parking lot cowboys is much more efficient according to industry expert sources....

 

[quogueelectric]

I am not smart enough to narow the questions down.

In my home office right now I have 10A of load on a 14ga wire. When I hit the print button on the computer, the lights dim, and the computer battery picks up, and the printer resets. The printer is 6 amps. I am 30 feet from the electrical panel. This is just my example of a real world situation. Not all people have this kind of load in a room. What I am trying to pry out of everyones brain is does/will/could this kind of situation, lead to premature failure of electrical equipment? I could list lots of examples of light load and heavy load situations on branch circuits. That is not what I want. I want to know if an average branch circuit is more efficient in 14 or 12. I guess it depends on your average....Average TV, Lights and a Vacume will crate this situation.

Thank you

Usually when lights dim upon a load pickup as described usually indicates to me that the local pole mount transformer is overloaded. Check to see how many houses are on this pole pig with all of the ac units humming in florida a 25k xformer can get loaded pretty quick.

 

[quogueelectric]

090919-1151 EST

Palmbay:

Is your question really about efficiency or something else?

First, consider efficiency in the sense of the ratio of useful power to input power. Also assume a constant resistance load of 10 ohms and a constant source voltage of 120 V. Four calculations, 50 ft and 100 ft for the
distances from the source voltage, and #14 (2.525 ohms/1000 ft) and #12 (1.588 ohms/1000 ft).

Disregard any slight error in my least significant digit. Also check my math for any mistakes.

50 ft distance and #14.
Copper resistance = 0.2525 ohms. Current = 120/10.2525 = 11.7045 . Power dissipated in the #14 = 34.59 W or 0.69 W per foot. Power dissipated in the load 10.2525*10.2525*10 = 1369.94 W. Input power is 120*11.7045 = 1404.54 W. Efficiency = 1369.94/1404.54 = 0.9754 or 97.54%.

100 ft distance and #14.
Copper resistance = 0.5050 ohms. Current = 120/10.5050 = 11.4231 . Power dissipated in the #14 = 65.896 W or 0.659 W per foot. Power dissipated in the load 11.4231*11.4231*10 = 1304.88 W. Input power is 120*11.4231 = 1370.78 W. Efficiency = 1304.88/1370.78 = 0.9519 or 95.19%.

So we went from a loss of 2.46% to 4.81%. Approximately doubling the losses, and reduced the output power by 2.4% .

If you had a motor load that was constant power rather than constant resistance, then loss percentage would be greater for the longer run.

50 ft #12.
Copper resistance = 0.1588 ohms. Current = 120/10.1588 = 11.8124 . Power dissipated in the #12 = 22.16 W or 0.44 W per foot. Power dissipated in the load 10.2525*10.2525*10 = 1395.33 W. Input power is 120*11.8124 = 1417.49 W. Efficiency = 1395.33/1417.49 = 0.9844 or 98.44%.

100 ft #12.
Copper resistance = 0.3176 ohms. Current = 120/10.2525 = 11.6306 . Power dissipated in the #12 = 42.96 W or 0.43 W per foot. Power dissipated in the load 10.2525*10.2525*10 = 1352.71 W. Input power is 120*11.7045 = 1395.67 W. Efficiency = 1352.71/1395.67 = 0.9692 or 96.92%.

Loss changed from of 1.66% to 3.08%. Approximately doubling the losses, and reduced the output power by 1.54% .


Consider a continuous load, then the losses in the #14 wire at 100 ft per year are 0.066*8760 = 578 KWH or about 578*0.13 = $75/year.

For #12 wire I get about 377 KWH or $49/year.

This kind of load might not be an unreasonable average load for a typical home. So is saving $25 per year on power loss worth the extra cost of the copper wire and installation. This is not a complete story, and 100 ft is probably long for an average distance for an average home. The 10 ohm continuous load is probably a good average load for a lived in home. That is an average consumption of 1.44 KWH.


Maybe more important is a question of voltage drop. At 100 ft, #12, and 10 ohm load the drop is about 0.3176/10.3176 = 3.1 % or 3.69 V. Change to #14 and the result is about 5.8 V. Both of these will cause noticeable flicker on an incandescent lamp.

Consider your computer problem. You need to make some voltage measurements. Also do some controlled voltage change measurements with the on-off of a 1500 W heater. You have a low source voltage, maybe 110 or lower (mine is about 122 to 125), and/or high resistance connections in you path from the pole transformer to your computer.

I have a UPS that trips into backup at about 100 V, then produces about 111 V, returns to non-backup at about 107 V. In other words the hysteresis is about 7 V. From memory I had one that went into backup at about 105 V or somewhat higher. That one could be a real nuisance in an area with a nominal 110 supply.

Today my voltage is about 123.8 V at my work bench and adding the 1500 W load it drops to 119.9 or a 3.9 V drop for about 12 A. This same load produces about 0.8 V change at my main panel, and most of this is the source impedance of the pole transformer.

At a more normal outlet in my house, than the bench, I get about a 1.4 V drop of which nearly 0.8 is the pole transformer drop. I have no circuit that has less than 2 breakers in series and the main fuses. The fuses have virtually no drop and the breakers are in the 0.1 V range.

You need to look at where the voltage drops are in your circuit that supplies your computer and printer.

I have no simple answer of how to explain whether to use #12 or #14 other than to say --- #12 for any receptacle and use 20 A breakers. Use #14 and 15 A for light circuits.

I am not in the business of wiring homes or businesses. So my criteria is quite different than for a typical home. For my own home the rough criteria are a main panel that supplies the 240 stuff, subpanels, a 120 circuit to outlets near each subpanel, several nearby 120 circuits including a separate 120 for each furnace. Virtually all lights and other outlets come from the subpanels. All 120 V receptacles are #12, and all lights are #14. All lights are from GE low voltage relays and wired with #14. Each bank of 3 relays is from a 15 A breaker. Every major room has two or more circuits for receptacles.

I believe your biggest problem in general is voltage drop. You could probably use #14 in most cases without great problems. Don't daisy chain thru receptacles and don't use any back stab (press in connections).

If you need to supply a 1.5 HP DeWalt radial arm saw use 240, and you could probably still use #14 over 25 to 50 ft. The same saw at 120 V and 100 ft of #12 is real marginal. These are voltage drop problems because of the high inrush current with a large diameter blade and its inertia. If I remember correctly, at the end of about 100 ft of #12 and a 120 circuit the inrush is about 70 to 80 A and lasts for several seconds.

For a kitchen it is probably better to have more #14 circuits than fewer #12. Obviously code determines what you do in this respect. Most kitchen devices would be 1500 W or less. The kitchen is probably a current limit problem rather than voltage drop. Thus, one circuit per device makes sense. You design for what the application may be and what code requires. You might very well design with more circuits than what code requires. My house was built 43 years ago and I have a separate 20 A circuit for the dishwasher, disposal, and four more circuits in the counter area. Lights are obviously separate based on my above discussion.

.

That was a beautiful answer I am floored by your effort.

 

[neutral]

Well duh. :grin:

Is that your final answer?