Voltage Drop Calculation

The method

Every one of these examples runs the same five steps. Only the multiplier and the source of the impedance change.

1. Find the conductor impedance per 1000 ft (or per km) for the size, material and temperature.
2. Reduce the run to a one-way length in the unit the impedance is quoted per.
3. Apply the multiplier: 2 for DC and single-phase, √3 for balanced three-phase.
4. Multiply by the current to get volts.
5. Divide by the source voltage to get a percentage.

1. DC: a 12 V solar circuit

A charge controller 25 ft from a battery bank, carrying 15 A on 10 AWG stranded copper. How much voltage arrives?

Step 1: resistance

10 AWG stranded copper is 1.24 Ω per 1000 ft at 75 °C, from NEC Chapter 9, Table 8.

Step 2: length

L = 25 ÷ 1000 = 0.025 thousands of feet.

Steps 3 and 4: multiplier and current

DC is a two-wire circuit, so k = 2:

V_drop = 2 × 15 A × 1.24 Ω/1000 ft × 0.025
V_drop = 0.93 V

Step 5: percentage

%V_drop = (0.93 ÷ 12) × 100 = 7.75%
V_load  = 12 − 0.93 = 11.07 V

2. Single-phase: a 120 V branch circuit

A 20 A load 100 ft from the panel on 12 AWG copper. This is the textbook case, and it fails the usual target.

Step 1: resistance

12 AWG stranded copper: 1.98 Ω per 1000 ft at 75 °C.

Steps 2 to 4

L = 100 ÷ 1000 = 0.1
V_drop = 2 × 20 A × 1.98 Ω/1000 ft × 0.1
V_drop = 7.92 V

Step 5

%V_drop = (7.92 ÷ 120) × 100 = 6.60%
V_load  = 120 − 7.92 = 112.08 V

What to do about it

12 AWG copper is rated 25 A at the 75 °C column, so ampacity is not the problem. Voltage drop is. Stepping to 8 AWG gives:

V_drop = 2 × 20 × 0.778 × 0.1 = 3.11 V  →  2.59%

Two sizes up, and the run passes. This is the pattern on 120 V circuits: ampacity stops binding early and voltage drop takes over.

3. Three-phase: a 480 V feeder with power factor

A 150 A motor load 350 ft from the switchgear on 250 kcmil copper in steel conduit, at 0.85 power factor. This is where a resistance-only calculation starts to mislead.

Step 1: effective impedance

From NEC Chapter 9, Table 9 for 250 kcmil uncoated copper in steel conduit: AC resistance R = 0.054 Ω and reactance X_L = 0.052 Ω per 1000 ft.

θ      = arccos(0.85) = 31.79°
sin θ  = 0.5268
Z_e    = R cos θ + X_L sin θ
Z_e    = 0.054 × 0.85 + 0.052 × 0.5268
Z_e    = 0.0459 + 0.0274 = 0.0733 Ω per 1000 ft

Steps 2 to 4

Balanced three-phase, so k = √3:

L = 350 ÷ 1000 = 0.35
V_drop = 1.732 × 150 A × 0.0733 Ω/1000 ft × 0.35
V_drop = 6.66 V  (line to line)

Step 5

%V_drop = (6.66 ÷ 480) × 100 = 1.39%
V_load  = 480 − 6.66 = 473.34 V

What resistance alone would have said

V_drop = 1.732 × 150 × 0.054 × 0.35 = 4.91 V  →  1.02%

A 36% understatement. At 1.39% against 1.02% both still pass a 3% target, so nothing goes wrong here, but push the run to 750 ft and the resistance-only figure says 2.2% while the real answer is 3.0%. That is the difference between passing and failing.

4. Metric: a 230 V cable run

A 25 A single-phase load 40 m from the distribution board on 6 mm² copper.

Step 1: resistance from resistivity

Copper at 75 °C has a resistivity of about 0.0209 Ω·mm²/m. For 6 mm²:

R = 0.0209 ÷ 6 = 0.003483 Ω per metre
  = 3.483 Ω per km

Steps 2 to 4

V_drop = 2 × 25 A × 0.003483 Ω/m × 40 m
V_drop = 6.97 V

Or straight from the resistivity form of the formula:

V_drop = (2 × ρ × L × I) ÷ A
V_drop = (2 × 0.0209 × 40 × 25) ÷ 6 = 6.97 V

Step 5

%V_drop = (6.97 ÷ 230) × 100 = 3.03%
V_load  = 230 − 6.97 = 223.03 V

5. Resistor: Ohm’s law on its own

A different question with the same physics. An LED needs 20 mA and drops 2.1 V; the supply is 12 V. What does the series resistor drop, and how much power must it shed?

Step 1: the voltage the resistor has to absorb

V_resistor = 12 − 2.1 = 9.9 V

Step 2: the resistance that produces that at 20 mA

R = V ÷ I = 9.9 ÷ 0.02 = 495 Ω

The nearest standard value is 470 Ω. Recompute with the real part:

I = 9.9 ÷ 470 = 0.02106 A = 21.06 mA

Step 3: power

P = V × I = 9.9 × 0.02106 = 0.209 W

A quarter-watt resistor would be running at 84% of its rating: hot, and drifting. A half-watt part is the right choice.

Notice what is absent: no length, no material, no multiplier. The resistor calculator covers this case and the series-divider version of it.

Where these go wrong

Entering a round-trip length as one-way

The commonest error by a distance. The multiplier already accounts for the return conductor. If you measured 200 ft of cable to serve a load 100 ft away, the one-way figure is 100 ft.

Getting the units of L wrong

Resistance in Ω per 1000 ft needs L in thousands of feet. Entering 250 instead of 0.25 gives an answer a thousand times too large, obvious enough to catch, unlike the factor-of-two error above.

Choosing single-phase for a three-phase circuit

Using 2 where √3 belongs overstates the drop by 15%, which makes you buy conductor you do not need. Using √3 where 2 belongs understates it, which is worse.

Ignoring reactance on a big inductive feeder

Fine at unity power factor and on small conductors. On large conductors at low power factor in steel conduit it understates the drop by a third or more, as example 3 shows.

Assuming a passing drop means a compliant conductor

Voltage drop and ampacity are separate checks and a conductor has to satisfy both. See the wire size calculator, which reports each one separately.

Questions

To run any of these with your own numbers and see the substitutions written out as you type, use the voltage drop calculator. For the formulas behind each step, see the voltage drop formula.