EngineeringFormula

Electrical

How to Calculate Voltage Drop

Voltage drop is the voltage lost in a conductor because the wire has resistance. The longer the run, the higher the current, or the smaller the conductor area, the more voltage is lost before the load receives power.

Final answer from the example

Voltage drop

6.90 V

Lost in the conductor run

Percent drop

2.87%

Compared with a 240 V supply

Load voltage

233.1 V

Estimated voltage at the load

Key formulas

Single-phase or DC voltage drop

Vdrop = 2 × I × ρ × L / A

The factor 2 accounts for the out-and-back current path in a two-wire circuit. L is the one-way run length.

Three-phase voltage drop

Vdrop = √3 × I × ρ × L / A

Balanced three-phase circuits use √3 instead of 2 because of the phase relationship between conductors.

Percent voltage drop

Percent drop = (Vdrop ÷ Supply voltage) × 100

Percent drop is the easiest way to compare circuits at different supply voltages.

Load voltage

Load voltage = Supply voltage - Vdrop

This estimates the voltage available at the far end of the conductor run under the stated load.

Variables and units

SymbolMeaningTypical units
VdropVoltage lost in the conductor runV
ILoad current carried by the conductorA
ρConductor resistivity at the reference temperatureΩ·m
LOne-way conductor length from source to loadm or ft
AMetal cross-sectional area of the conductorm², mm², kcmil
VNominal supply voltageV

Quick reference conversions

Single-phase/DC multiplier

2

Accounts for the out-and-back conductor path.

Three-phase multiplier

√3

Use for balanced three-phase voltage drop.

Copper resistivity

1.7241×10⁻⁸ Ω·m

Reference value at 20°C.

Aluminum resistivity

2.826×10⁻⁸ Ω·m

Higher than copper, so drop is larger for the same area.

Branch or feeder target

3%

Common NEC informational-note design recommendation.

Combined target

5%

Common feeder plus branch total design recommendation.

Step-by-step solved example

Example problem

Estimate voltage drop for a 240 V single-phase copper branch circuit carrying 10 A over a 50 m one-way run using 2.5 mm² copper conductors. Use copper resistivity ρ = 1.7241×10⁻⁸ Ω·m at 20°C.

1. Choose the correct circuit multiplier

This is a single-phase circuit, so use k = 2. A balanced three-phase circuit would use k = √3 instead.

2. Convert conductor area to square metres

The formula uses SI base units. 2.5 mm² = 2.5×10⁻⁶ m² because 1 mm² = 10⁻⁶ m².

3. Substitute values into the voltage drop formula

Vdrop = (2 × 10 A × 1.7241×10⁻⁸ Ω·m × 50 m) ÷ 2.5×10⁻⁶ m².

4. Calculate volts dropped

Vdrop = 6.90 V. That means about 6.90 volts are lost in the copper run at the stated load current.

5. Convert to percent drop

Percent drop = (6.90 V ÷ 240 V) × 100 = 2.87%. The estimated load voltage is 240 V - 6.90 V = 233.1 V.

6. Interpret the result

A 2.87% drop is below the common 3% design target for a single branch circuit or feeder. If the result were above 3%, a larger conductor, shorter run, or lower current would be worth reviewing.

Practical field notes

Use one-way length, not round-trip length

The single-phase formula already multiplies by 2 for the return path. If you enter round-trip length and also use the factor 2, the result will be doubled.

Voltage drop is a load-current calculation

A circuit with no load has essentially no measured voltage drop. Use realistic operating current when estimating voltage at the load.

Voltage drop is not the same as ampacity

A conductor can be large enough for ampacity but still have too much voltage drop on a long run. Always satisfy code ampacity, overcurrent protection, temperature limits, and voltage drop separately.

The simple resistivity formula is a reference estimate

For long AC feeders, larger conductors, conduit effects, or low power factor loads, impedance and reactance can matter. Detailed design may need NEC/IEC table methods or engineering review.

Common mistakes to avoid

  • Entering total round-trip length instead of one-way length while also using the single-phase factor of 2.
  • Using mm² directly as m² without converting 2.5 mm² to 2.5×10⁻⁶ m².
  • Comparing volts dropped without converting to percent of supply voltage.
  • Treating the NEC 3%/5% voltage-drop guidance as the only conductor-sizing requirement.
  • Ignoring conductor temperature, aluminum vs copper material, or three-phase vs single-phase circuit type.

When to use the calculator instead

Use the calculator when you want to compare copper vs aluminum, single-phase vs three-phase, different conductor areas, or percent drop against a supply voltage. It keeps length, area, resistivity, and percent conversions consistent.

Calculation FAQs

What is voltage drop?

Voltage drop is the voltage lost as current flows through conductor resistance. It increases with higher current, longer wire length, and smaller conductor area.

Why does single-phase use 2 in the voltage drop formula?

A single-phase or DC two-wire circuit has an outgoing conductor and a return conductor, so the current path is effectively twice the one-way distance.

Why does three-phase use √3?

For a balanced three-phase circuit, the phase relationship between conductors gives the line-to-line voltage drop multiplier √3 rather than 2.

Is 3% voltage drop a legal code limit?

In the NEC, the common 3% branch or feeder and 5% combined values are informational-note design recommendations in most cases, not standalone mandatory rules. Local requirements and special equipment rules can be stricter.

How do I reduce voltage drop?

Use a larger conductor, shorten the run, reduce current where practical, use a higher distribution voltage when design allows, or split loads across circuits.

References

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