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Voltage Drop in 12 V and 24 V DC Solar and Battery Cables

At 12 V or 24 V the same power needs ten to twenty times the current of a 230 V circuit, so a cable that would be generous on mains can lose several percent of the voltage in a few metres. In a battery system that lost voltage is lost energy and can make an inverter trip on low voltage while the battery is still well charged.

The calculator below is preset for a 24 V circuit carrying 40 A over a 6 m one-way route in 16 mm² copper at a conductor temperature of 50 °C, a typical battery-to-inverter or controller-to-battery link. Change the current, length and size to your own run; the result shows the drop in volts and percent and the smallest size that keeps within the limit.

Last reviewed

Current through the cable's resistance and reactance loses voltage along the run: the load sees V₂ = V₁ − ΔV, and ΔV grows in proportion to length and current.

For estimation and education. Not a substitute for a licensed engineer, the applicable codes (IEC 60364, NEC/NFPA 70, BS 7671, local regulations) or manufacturer data. Read the disclaimer.

Inputs

Circuit

Conductor

Material
Size system
°C

70 °C PVC, 90 °C XLPE at full load

Ω/km

From IEC 60228 or maker data; overrides resistivity

Circuit

A
m
Length unit
V
%

Common design limits: 3 % lighting, 5 % other

Result

Voltage drop — Engineerly. Estimate for education only; verify against applicable codes and manufacturer data.

Voltage drop
0.5782V
Voltage drop
2.41%

Within your 3 % limit.

Voltage at load end
23.422V
Conductor resistance at temperature
1.205mΩ/m
Route length (one way)
6 m
Smallest IEC size meeting 3 %
16 mm² (2.41 %)

Working, with your numbers

  1. 1. Resistance at conductor temperature
    R_T = ρ20 (1 + α (T − 20)) / A
    R_T = 0.017241 × (1 + 0.00393 × (50 − 20)) / 16 mm²
    = 1.2046 mΩ/m
  2. 2. Voltage drop (DC, out and back)
    ΔV = 2 × I × L × R / n
    ΔV = 2 × 40 A × 6 m × 0.00120461 Ω/m / 1
    = 0.5782 V
  3. 3. Percentage of nominal voltage
    ΔV% = ΔV / V × 100
    ΔV% = 0.578211 V / 24 V × 100
    = 2.41 %

Assumptions

  • Copper ρ20 = 0.017241 Ω·mm²/m, α = 0.00393 /K, conductor at 50 °C.
  • Resistance from resistivity and nominal area (DC resistance; slightly below tabulated maximum values for stranded conductors).
  • DC two-wire circuit: out and back, no reactance.
  • Load current is constant along the route (a single load at the far end).
  • The approximate formula ΔV = k·I·L·(R cos φ + X sin φ) is accurate for drops of a few percent.
System: DC
Conductor: Copper 16 mm² × 1
Temperature: 50 °C
Current: 40 A
Length: 6 m
Voltage: 24 V
Limit: 3 %

Check before you rely on this

  • Voltage drop is only one check. The conductor must also carry the current (IEC 60364-5-52 / NEC 310.16 ampacity with grouping and ambient corrections) and satisfy fault-disconnection and short-circuit withstand requirements.
  • Estimation and education only. Not a substitute for a licensed engineer, the applicable code (IEC 60364, NEC/NFPA 70, BS 7671 or local regulations) or manufacturer data.

Notes for this case

  • DC drop is computed out and back (factor 2) with no reactance.
  • Many installers aim for 1–3 % on DC battery and PV cables; follow your inverter and controller manuals, which often state a maximum cable length per size.
  • Ampacity and fusing still apply: a battery cable must be protected by a DC-rated fuse or breaker close to the battery.

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How these formulas were checked: methodology and testing.