How do I size a wire? From the load to the voltage drop, step by step

Sizing a wire takes three linked calculations. First, the load’s watts divided by the voltage give the current. Second, the wire’s length, gauge, and metal give its resistance. Third, current times resistance is the voltage lost in the wire. If that drop is over about 3 percent, choose a thicker gauge or a shorter run.

3 steps · 2 links. Change any value that isn’t linked. Everything after it updates.

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Step 1 · LIN-0003
How many amps does it draw?
Current
12.5 A
Power
1.5 kW
Supply voltage
120 V
Appliance power (P)
On the label or the nameplate, in watts.
Supply voltage (V)
More settings
Breaker rating
Continuous loads should stay under 80% of this.
Show the work
1. Power = voltage × currentP = VI ;\Rightarrow; I = \frac{P}{V}
2. With your numbersI = \frac{1500\,\mathrm{W}}{120\,\mathrm{V}} = 12.5\,\mathrm{A}
3. 80% breaker check0.8 \times 15\,\mathrm{A} = 12\,\mathrm{A}
Step 2 · RAT-0002
Wire resistance: longer and thinner resist more
Wire resistance
0.414 Ω
Total conductor
50 m
Cross-section
2.08×10^-6 m²
Cable run (one way) (L)
Wire gauge
More settings
Metal
Count out and back (both wires)
Show the work
1. Resistance of a wireR = \frac{\rho L}{A}
2. Area of 14 AWGA = \frac{\pi d^2}{4} = \frac{\pi (1.628\,\mathrm{mm})^2}{4} = 2.081\,\mathrm{mm^2}
3. With your numbersR = \frac{1.724\times 10^{-8}\,\Omega\cdot\mathrm{m} \times 50\,\mathrm{m}}{2.081\times 10^{-6}\,\mathrm{m^2}} = 0.4142\,\Omega
from step 1: 12.5 A →from step 2: 0.414 Ω →
Step 3 · LIN-0004
Voltage drop: what the wire takes before the load gets it
Voltage drop
5.18 V
Drop (% of supply)
4.32%
Power lost as heat
64.7 W
Voltage at the load
115 V
Current (I)12.5 A
from LIN-0003 · Current ·
Wire resistance (both conductors) (R)0.414 Ω
from RAT-0002 · Wire resistance ·
Supply voltage (V)
Show the work
1. Ohm’s law on the wire\Delta V = IR = 12.5 \times 0.4142 = 5.178\,\mathrm{V}
2. As a percent of supply\frac{5.178}{120} \times 100 = 4.315\%
3. Heat in the wireP = I^2R = (12.5)^2 \times 0.4142 = 64.73\,\mathrm{W}

Final results

Voltage drop (LIN-0004): 5.18 V
Drop (% of supply) (LIN-0004): 4.32%
Power lost as heat (LIN-0004): 64.7 W
Voltage at the load (LIN-0004): 115 V

Step by step

Find the current the appliance draws: I = P / V.
Find the resistance of the wire run, counting both conductors: R = ρL / A.
Multiply them to get the voltage drop: ΔV = I × R. Compare it with the supply voltage.
If the drop is over 3%, try the next thicker gauge (a smaller AWG number) and watch every step update.

Each step is a full exhibit: How many amps does it draw?, Wire resistance: longer and thinner resist more, Voltage drop: what the wire takes before the load gets it.

Questions people ask

How much voltage drop is acceptable?

A common guideline is 3% on a branch circuit and 5% total. Motors and electronics may need less.

Why does the gauge change the voltage drop?

A thicker wire has more cross-section, so less resistance. Less resistance means less voltage lost for the same current.

Can I change the steps?

Yes. Change the load, the length, or the gauge and every later step recalculates. Open it as a board to add or remove links.