Direct current
For battery banks, solar arrays, vehicles and low-voltage lighting, where the supply is small enough that half a volt matters. Enter the current, the conductor and the run, and see what is left at the load.
7.75% is above the 3% limit you set. About 7.75% of the 12 V supply is lost in the conductors under these conditions, leaving 11.07 V at the load. That is 0.93 V dropped across the run, out and back.
Your voltage drop appears here
Enter a supply voltage, a load current and a cable length. The result updates as you type.
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7.75%
0.93 V lost against a3% limit
7.75% is above the 3% limit you set. About 7.75% of the 12 V supply is lost in the conductors under these conditions, leaving 11.07 V at the load. That is 0.93 V dropped across the run, out and back.
To bring it down: increase the conductor size, shorten the run, split the load across parallel conductors, or supply the load at a higher voltage. Doubling the conductor area roughly halves the drop; halving the run length halves it exactly.
Conductor resistance per 1000 ft
R = table value for size and material10 AWG copper, stranded, at 75 °CResult: 1.24 Ω per 1000 ft
From NEC Chapter 9 Table 8, direct-current resistance at 75 °C.
Voltage drop
V_drop = 2 × I × R × LV_drop = 2 × 15 A × 1.24 Ω/1000 ft × 0.0250 (1000 ft)Result: 0.93 V
Current flows out along one conductor and back along the other, so it travels twice the one-way run length.
Voltage drop percentage
%V_drop = (V_drop ÷ V_source) × 100%V_drop = (0.93 V ÷ 12 V) × 100Result: 7.75%
Voltage at the load
V_load = V_source − V_dropV_load = 12 V − 0.93 VResult: 11.07 V
Power lost in the conductors
P_loss = I² × R_path × number of current-carrying pathsP_loss = 15² × 0.031 Ω × 2Result: 14.0 W
Only resistance dissipates power. Reactance stores and returns energy each cycle rather than turning it into heat.
Resistance: NEC Chapter 9, Table 8 (DC resistance at 75 °C)
Formula
There is no reactance term and no power factor. In a DC circuit the only thing opposing current is resistance, so this is an exact calculation rather than an approximation. The accuracy of the answer depends entirely on how well you know the current, the length and the conductor temperature.
Reference
Copper conductors, one-way run length, resistance at 75 °C. Anything over 3% is marked.
| Size | 10 ft | 25 ft | 50 ft | 100 ft |
|---|---|---|---|---|
| 14 AWG | 7.85%, over 3% | 19.63%, over 3% | 39.25%, over 3% | 78.50%, over 3% |
| 12 AWG | 4.95%, over 3% | 12.38%, over 3% | 24.75%, over 3% | 49.50%, over 3% |
| 10 AWG | 3.10%, over 3% | 7.75%, over 3% | 15.50%, over 3% | 31.00%, over 3% |
| 8 AWG | 1.94% | 4.86%, over 3% | 9.73%, over 3% | 19.45%, over 3% |
| 6 AWG | 1.23% | 3.07%, over 3% | 6.14%, over 3% | 12.28%, over 3% |
| 4 AWG | 0.77% | 1.93% | 3.85%, over 3% | 7.70%, over 3% |
| 2 AWG | 0.49% | 1.21% | 2.43% | 4.85%, over 3% |
Computed with the site calculator from NEC Chapter 9, Table 8 resistance values at 75 °C.
Read down a column to see how much conductor a given distance costs you, or across a row to see how far one size will reach. Halving the current halves every figure; doubling the supply to 24 V halves them again. For an AC circuit, or one where reactance and power factor come into play, the main voltage drop calculator covers every system type.
Questions
Because the percentage is what counts, and the same volts lost are a far bigger share of a smaller supply. Losing 0.5 V on a 120 V circuit is 0.42% and nothing notices. Losing 0.5 V on a 12 V circuit is 4.2%, which is enough to dim lighting noticeably and to stop some equipment working. Low-voltage systems also draw much higher current for the same power, and drop rises directly with current, so the two effects compound.
The multiplier is the same (2, because current goes out on one conductor and back on the other), but DC has no reactance and no power factor. Voltage drop = 2 × I × R × L, using plain conductor resistance. That makes DC the simplest case and also the one where a resistance-only calculation is exactly right rather than an approximation.
There is no code figure for this the way there is for premises wiring. 3% is the usual working target for a main run and many solar installers design to 2% or tighter between panels and charge controller, because that loss is energy you paid for and never get back. Equipment data sheets often set the real constraint: check the minimum input voltage of whatever is at the far end.
Measure the distance from the source to the load. That is the one-way length, and it is what the formula expects. The 2 in the formula already accounts for the return conductor. If you know the total length of cable you pulled instead, switch the calculator to round-trip and enter that; it halves the figure for you.
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