Voltage Drop Calculator for 12 V and 24 V: How much voltage is lost in the wire?

Over 16 ft (5 m) of 6 mm² (≈ 10 AWG) wire, almost 0.6 V is lost at 20 A – around 5% of the system voltage, enough for an inverter to report low voltage or a charger to stop charging properly. This calculator shows how much voltage and power is lost in your wire, and whether the gauge is sufficient for the run.

Voltage drop

0.581 V

4.5% of the system voltage

Voltage drop (recommendation ≤ 3%)5 / 3 %

Voltage at the load: 12.22 V
Power loss in the wire: 11.6 W
Wire resistance (out + return): 29 mΩ
Current: 20 A

Over 3% voltage drop is considered the upper limit of the common recommendation in a 12 V system; for charging circuits (solar, DC-DC charger) 1–2% is sensible.
Ampacity (heating) is a separate requirement: the gauge must additionally be rated for the current per the manufacturer's/code's wire table, and the wire must be fused accordingly.

Show calculation

1Wire resistance: 2 × 16 ft ÷ (56 × 9 AWG (6 mm²)) = 0.029 Ω
2Voltage drop: 20 A × 0.029 Ω = 0.581 V (4.5% of 12.8 V)
3Power loss: 0.581 V × 20 A = 11.6 W

How it's calculated

Every wire has a resistance that depends on length, cross-section and material. For DC the length counts twice, because the current flows out and back.

R [Ω] = 2 × L [m] ÷ (κ × A [mm²])
ΔU [V] = I [A] × R

κ (copper, 68 °F / 20 °C) = 56 m/(Ω·mm²). Example: 2 × 5 m ÷ (56 × 6 mm²) = 0.0298 Ω; 20 A × 0.0298 Ω = 0.595 V

The power loss P = ΔU × I is turned into heat in the wire – at high currents that quickly reaches 10 W and more.

How much voltage drop is acceptable?

Up to 3% is the common recommendation for load circuits in a 12 V system.
1–2% for charging circuits (solar controller → battery, DC-DC charger → battery), so the charge voltage actually arrives.
Over 5% causes malfunctions: inverter low-voltage shutdown, flickering lights, weak pumps.

Voltage drop is only one of two criteria. Ampacity (permissible heating) depends on the wire type, installation method and ambient temperature and must be checked against manufacturer tables or codes (e.g. ABYC E-11 for boats, NEC) – as must the fusing.

Worked example

A DC-DC charger delivers 50 A over 13 ft (4 m) of wire (one-way length) with 16 mm² (≈ 6 AWG) into a 12.8 V LiFePO4 battery.

Inputs

System voltage: 12 V (LiFePO4 12.8 V)
Specify current as: Current in A
Current: 50 A
One-way wire length: 13.1 ft
Conductor cross-section: 16 mm² (≈ 6 AWG)
Conductor temperature (copper conductivity): 68 °F / 20 °C – κ = 56 m/(Ω·mm²)

Result

0.446 V

Voltage drop

Voltage at the load: 12.35 V
Power loss in the wire: 22.3 W
Wire resistance (out + return): 8.9 mΩ
Current: 50 A

Calculation

Wire resistance: 2 × 13.1 ft ÷ (56 × 5 AWG (16 mm²)) = 0.0089 Ω
Voltage drop: 50 A × 0.0089 Ω = 0.446 V (3.5% of 12.8 V)
Power loss: 0.446 V × 50 A = 22.3 W

The variables explained

One-way wire length: Distance from the source to the load; the return conductor is included automatically (factor 2).
Current (A): Continuous current flowing through the wire. From power: I = P ÷ U.
Conductor cross-section (mm²): Metric standard sizes per IEC 60228: 1.5 · 2.5 · 4 · 6 · 10 · 16 · 25 · 35 · 50 · 70 · 95 mm²; AWG equivalents are shown next to each size.
Conductivity κ: Copper 56 m/(Ω·mm²) at 68 °F (20 °C), about 48 at 158 °F (70 °C). Aluminum only about 35.

Common mistakes

Calculating only the one-way length: the current flows out and back – the wire length counts twice.
Mixing up AWG and mm²: 6 AWG ≈ 13.3 mm², 4 AWG ≈ 21.2 mm², 2 AWG ≈ 33.6 mm².
Ignoring contact resistance: poor crimps, corroded terminals and fuse holders often cause more drop than the wire itself.
Forgetting the drop on extensions (solar cable across the roof) – long thin wires cost charging power.

Assumptions and limits

Pure copper conductor (no CCA/aluminum wire), conductivity 56 or 48 m/(Ω·mm²).
Contact resistance at terminals, fuses and connectors is not included.
Ampacity and fusing are not assessed – no installation approval.