12V Wire Size Calculator

For 12V DC circuits, size wire by voltage drop, not ampacity. Low system voltage means even a small voltage loss is a large percentage, so a 0.36 V drop is already 3% on a 12 V system. Using the circular-mil formula CM = (2 × length × amps × 10.75) ÷ allowed volts, a 20 A load over 10 feet at a 3% target needs about 11,944 circular mils — which maps to 8 AWG copper.

This 12V wire size calculator is built for marine, RV, solar, and automotive DC work, where the American Boat and Yacht Council (ABYC) and RV standards demand tight voltage-drop limits (3% for critical circuits, 10% for non-critical). Enter your current, one-way length, target drop, and 12 V or 24 V system voltage, and it returns the AWG plus the actual drop you will see.

DC wire sizing uses circular mils (CM) — the cross-sectional area of the conductor — because area is what determines resistance and therefore voltage drop. The standard marine/RV equation is:

Where:

• L = one-way circuit length in feet (the 2 accounts for the round trip).
• I = current in amps.
• K = 10.75, the resistivity constant for copper in circular-mil-ohms per foot. (Aluminum uses ~17.7.)
• Vdrop = the allowable voltage drop in volts = system voltage × target percentage.

Once you have the required circular mils, you pick the smallest AWG whose area meets it. A circular mil is the area of a circle 0.001 inch in diameter; 8 AWG is 16,510 CM, 6 AWG is 26,240 CM, 4 AWG is 41,740 CM. Because each AWG step down roughly doubles the circular mils every three sizes, dropping voltage in half means going up about three gauge sizes. This is the same physics behind the general voltage drop calculator, expressed in the circular-mil form that DC/marine work prefers.

Example 1 — 20 A, 10 ft, 3%, 12 V: allowed drop = 12 × 0.03 = 0.36 V. Required CM = (2 × 10 × 20 × 10.75) ÷ 0.36 = 4,300 ÷ 0.36 = 11,944 CM. Smallest AWG ≥ 11,944 is 8 AWG (16,510 CM). Actual drop = 4,300 ÷ 16,510 = 0.26 V = 2.17%. Answer: 8 AWG.

Example 2 — 10 A LED run, 25 ft, 3%, 12 V: allowed = 0.36 V. Required CM = (2 × 25 × 10 × 10.75) ÷ 0.36 = 5,375 ÷ 0.36 = 14,931 CM → 8 AWG (16,510). Even a modest 10 A load over 25 feet needs surprisingly thick wire on 12 V — the penalty of low voltage.

Example 3 — same load on 24 V: 10 A, 25 ft, 3%, 24 V. Allowed = 24 × 0.03 = 0.72 V. Required CM = 5,375 ÷ 0.72 = 7,465 CM → 10 AWG. Doubling system voltage roughly halves the required copper — a core reason larger DC systems move to 24 V or 48 V.

Example 4 — 80 A windlass, 15 ft, 10%, 12 V: windlasses are non-critical, so ABYC allows 10%. Allowed = 1.2 V. Required CM = (2 × 15 × 80 × 10.75) ÷ 1.2 = 25,800 ÷ 1.2 = 21,500 CM → 6 AWG. At a 3% target it would need 4 AWG or larger — showing how the allowed-drop choice drives the gauge.

The same appliance wired at 12 V draws ten times the current it would at 120 V (Power = Volts × Amps). A 120 W load is 1 A at 120 V but 10 A at 12 V. Since voltage drop scales with current, low-voltage circuits lose far more voltage over the same distance — and because the budget (a 3% slice of 12 V is only 0.36 V) is tiny, the wire must be much larger.

This is why:

• A 12 V circuit may need 8 AWG where a 120 V version uses 14 AWG for the same power.
• Battery cables to inverters are often 2 AWG, 1/0, or 4/0 despite short lengths — the current is enormous.
• Solar and off-grid systems migrate to 24 V and 48 V as they grow, cutting current and copper cost dramatically.

The circular-mil method captures this automatically because it works in volts of allowable drop, not a fixed ampacity. For AC circuits and the standard 3%/5% branch-feeder rules, use the wire size calculator instead; for pure drop analysis at any voltage, the voltage drop calculator complements this tool.

Marine and RV wiring follow ABYC E-11 and RVIA guidance, which classify circuits by criticality:

• 3% maximum drop — critical circuits: navigation lights, bilge pumps, electronics, panelboard main feeds. Reliability and safety demand tight regulation.
• 10% maximum drop — non-critical circuits: cabin lighting, general accessories, windlasses, and other intermittent loads where a little dimming is acceptable.

The calculator defaults to 3% (the conservative, critical-circuit target) but lets you enter 10% for non-critical loads to avoid over-sizing. Marine wiring adds another requirement: stranded, tinned copper for corrosion resistance and vibration tolerance — solid wire is prohibited on boats. Marine-grade wire also carries slightly different ampacity ratings, but on 12 V systems voltage drop almost always governs before ampacity does, so the circular-mil result is the binding constraint.

Automotive DC (car audio, winches, auxiliary lighting) uses the same formula. High-current car-audio amplifiers are a classic case: a 100 A amp 15 feet from the battery needs 4 AWG or larger to avoid starving the amp under bass peaks.

Because voltage drop on 12 V circuits depends on both current and distance, a quick chart makes the pattern obvious before you reach for the calculator. The table below gives the smallest copper AWG that keeps a 12 V circuit under 3% drop, using the circular-mil method. Notice how the required gauge grows sharply with distance — the low-voltage penalty in action.

Read it as a starting point, then confirm your exact numbers above. The chart also explains why van and boat builders route heavy loads (inverters, refrigeration, windlasses) as close to the battery as possible: every foot of a 12 V run costs real copper. Where a load must sit far from the battery, raising the system to 24 V or 48 V, or relocating the fuse block nearer the load, is usually cheaper than the extra cable a long 12 V run demands.

1. Enter load current in amps. For DC devices rated in watts, divide watts by system voltage (e.g., 120 W ÷ 12 V = 10 A).
2. Enter one-way length from the source (battery or fuse block) to the load. The formula doubles it for the return path.
3. Set your target voltage drop. Use 3% for critical circuits and battery-to-inverter cables, or up to 10% for non-critical accessory loads.
4. Choose 12 V or 24 V. Watch the required copper shrink when you switch to 24 V — the same load needs roughly half the circular mils.

The tool returns the AWG, the actual drop percentage and volts for that gauge, and the required circular mils. If the actual drop is comfortably below your target, you have margin; if it sits right at the limit, consider the next size up for long-term reliability and future load growth.

Sizing 12V wire by voltage drop gives the conductor's minimum diameter, but two more checks complete a safe DC circuit: ampacity and overcurrent protection.

Ampacity still applies. Even though voltage drop usually governs, the chosen wire must also carry the current without overheating. On short, high-current runs — battery to inverter, battery to windlass — ampacity can actually be the binding limit. A 2,000 W inverter draws roughly 165 A at 12 V; even over a 5-foot run the cable must be 2/0 or larger to handle the current and the surge, regardless of what the drop formula alone suggests.

Fuse or breaker sizing. ABYC requires overcurrent protection within 7 inches of the battery positive terminal (or up to 72 inches if the conductor is in a protective sheath). The fuse protects the wire, so it must be rated at or below the conductor's ampacity — never higher. A common mistake is fitting an oversized fuse to "avoid nuisance trips," which defeats the protection and risks a cable fire. Coordinate the fuse to the wire, then verify the load draws less than the fuse rating.

Battery cable sizing. Main battery cables are a special case: they are short but carry the full system current, so they are sized for ampacity and mechanical robustness, typically 4 AWG, 2 AWG, 1/0, or 4/0 depending on the inverter. Use fine-stranded, tinned copper with properly crimped and sealed lugs. For the AC output side of an inverter, switch to the wire size calculator, and confirm overcurrent devices with the breaker size calculator.

• Sizing by ampacity alone. On 12 V, a wire that "handles the amps" often drops far too much voltage — always size by drop.
• Measuring one-way but forgetting the return. The formula's factor of 2 covers both conductors; enter only the one-way length.
• Using solid or non-tinned wire on boats. Marine circuits require stranded, tinned copper for vibration and corrosion resistance.
• Undersizing battery-to-inverter cables. A 1,000 W inverter pulls ~90 A at 12 V; short runs still need 4 AWG to 1/0.
• Ignoring fuse placement. Fuse each conductor near the battery per ABYC; wire size and fuse rating must be coordinated.

For AC-side sizing of an inverter's output, switch to the wire size calculator, and confirm overcurrent protection with the breaker size calculator.