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.
.