12 V Wire Gauge Calculator: What size wire does the circuit need?
A 1,500 W inverter draws about 140 A at 12 V – over 6.5 ft (2 m) of wire that needs 2 AWG (35 mm²) to keep the voltage drop under 3%. This calculator determines the calculated minimum cross-section and the next standard size for load and charging circuits in your RV.
Important: the result covers voltage drop. The wire's ampacity and the correct fuse must additionally be checked against manufacturer data and codes.
Result
Recommended minimum wire size
6 AWG (13.3 mm²)
calculated 8 AWG (8.5 mm²) at max. 3% voltage drop
- Calculated cross-section
- 8 AWG (8.5 mm²)
- Voltage drop with standard size
- 0.246 V
- Current
- 30 A
- Total conductor length (out + return)
- 20 ft
- The gauge from the voltage drop is a minimum. Ampacity (heating), installation method and fusing must additionally be checked against manufacturer data and codes – no installation approval.
Show calculation
- 1Permissible drop: 12.8 V × 3% = 0.384 V
- 2Cross-section: 2 × 10 ft × 30 A ÷ (56 × 0.384 V) = 8 AWG (8.5 mm²)
- 3Next standard size: 6 AWG (13.3 mm²) → actual drop 0.246 V (1.9%)
How it's calculated
The calculator solves the voltage drop formula for the cross-section. You specify how much voltage may be lost in the wire at most.
A [mm²] = 2 × L [m] × I [A] ÷ (κ × ΔU_max [V])The calculated value is rounded up to the next standard size. With US units these are the common AWG sizes 18 … 4/0 AWG; with metric units the IEC 60228 series 1.5 · 2.5 · 4 · 6 · 10 · 16 · 25 · 35 · 50 · 70 · 95 mm².
| Current | 6.5 ft (2 m) | 13 ft (4 m) | 20 ft (6 m) |
|---|---|---|---|
| 10 A | 14 AWG (1.9) | 12 AWG (3.7) | 10 AWG (5.6) |
| 30 A | 10 AWG (5.6) | 6 AWG (11.2) | 4 AWG (16.7) |
| 60 A | 6 AWG (11.2) | 4 AWG (22.3) | 2 AWG (33.5) |
| 120 A | 4 AWG (22.3) | 1/0 AWG (44.6) | 2/0 AWG (67.0) |
Worked example
A 2,000 W inverter (90% efficiency) is connected to a 12.8 V LiFePO4 with 5 ft (1.5 m) of wire; permissible voltage drop 2%.
Inputs
- System voltage: 12 V (LiFePO4 12.8 V)
- Specify current as: Power in W (e.g. inverter)
- Power: 2,000 W
- Efficiency (for inverters): 90 %
- One-way wire length: 4.92 ft
- Permissible voltage drop: 2 %
- Conductor temperature (copper conductivity): 68 °F / 20 °C – κ = 56 m/(Ω·mm²)
Result
1 AWG (42.41 mm²)
Recommended minimum wire size
- Calculated cross-section
- 2 AWG (36.33 mm²)
- Voltage drop with standard size
- 0.219 V
- Current
- 173.6 A
- Total conductor length (out + return)
- 9.8 ft
Calculation
- Current: 2,000 W ÷ (12.8 V × 90%) = 173.6 A
- Permissible drop: 12.8 V × 2% = 0.256 V
- Cross-section: 2 × 4.9 ft × 173.6 A ÷ (56 × 0.256 V) = 2 AWG (36.33 mm²)
- Next standard size: 1 AWG (42.4 mm²) → actual drop 0.219 V (1.7%)
The variables explained
- Continuous current (A)
- Maximum current flowing continuously. For inverters calculated from power, voltage and efficiency.
- One-way wire length
- Distance from the source to the load; the return conductor is included automatically.
- Permissible voltage drop (%)
- 3% for loads, 1–2% for charging circuits are common recommendations – not a vehicle code, but proven practice.
Common mistakes
- Sizing the wire only for voltage drop and forgetting ampacity: short runs with high current (inverter) can still get too hot at low drop if the wire is bundled or enclosed.
- Choosing the fuse for the load instead of the wire: the fuse protects the wire – it must match the gauge and sit close to the battery.
- Calculating cheap CCA wire (copper-clad aluminum) with copper values – it has about 35% less conductivity.
- Mixing up AWG and mm².
Assumptions and limits
- Copper conductor, conductivity 56 (68 °F / 20 °C) or 48 (158 °F / 70 °C) m/(Ω·mm²).
- Standard sizes up to 4/0 AWG (107 mm²) or 95 mm²; larger currents require parallel conductors or a 24 V system.
- No assessment of ampacity or fusing – manufacturer data and codes apply. Electrical installations should be checked by a professional.
Frequently asked questions
What size wire do I need for a 2,000 W inverter?
At 12.8 V and 90% efficiency about 175 A flows. For 5 ft (1.5 m) of wire and 2% voltage drop the calculator gives 36 mm² calculated, i.e. 2 AWG (33.6 mm²) isn't quite enough and 1/0 AWG (53.5 mm²) is the next common size; many manufacturers specify 2 AWG to 1/0 AWG for this power class. The inverter manufacturer's specification is binding.
Is 14 AWG (2.5 mm²) enough for a refrigerator circuit?
For 5 A over 20 ft (6 m) one-way length the result is 14 AWG at 3% drop. Compressor fridges are sensitive to low voltage – 12 AWG (4 mm²) is the safer choice on longer runs.