Electrical

DC Cable Voltage Drop: Formula, Calculation and Acceptable Limits

The DC voltage drop formula explained with the two-way circuit path, temperature-corrected resistance, worked PV and battery examples, and typical percentage limits.

Published by OneCalcApp Editorial TeamReviewed by Kodeeswaran Appavu 12 August 2026 10 min read

Voltage drop is the voltage lost in the conductors between the source and the load. In a DC circuit it is pure resistance — no reactance, no power factor — which makes it easy to calculate accurately and impossible to hide behind assumptions.

Run your own numbers in the DC Cable Sizing Calculator, which reports drop in volts, drop as a percentage, receiving-end voltage and I²R loss for every candidate cable size.

One-Way Length vs Two-Way Circuit Length

This is the single most common error in DC calculations.

Current leaves the source on the positive conductor and returns on the negative conductor. Both carry the full current, so the resistive path is twice the route length.

```text

One-way route length L = 100 m
Electrical circuit path 2 × L = 200 m

```

Enter the one-way length. A well-built calculator adds the return path internally. Entering 200 m for a 100 m run doubles the answer.

The Formula

$$\Delta V = I \times 2 R_T L$$

where $R_T$ is the conductor resistance in Ω/km at operating temperature and $L$ is the one-way length in km.

From resistivity:

$$R_{20} = \frac{\rho_{20} \times 1000}{A} \quad [\Omega/\text{km}]$$
$$R_T = R_{20}\,[1 + \alpha (T - 20)]$$
QuantityCopperAluminium
ρ₂₀ reference0.01724 Ω·mm²/m0.02826 Ω·mm²/m
ρ IEC 60364 Annex G normal service0.0225 Ω·mm²/m0.036 Ω·mm²/m
α0.00393 /K0.00403 /K

Where the manufacturer publishes a maximum DC resistance at 20 °C, use that value instead of computing one from resistivity.

Temperature-Corrected Resistance

Resistance rises with temperature, and cables do not operate at 20 °C.

Conductor temperatureCopper R relative to 20 °C
20 °C1.000
40 °C1.079
60 °C1.157
70 °C1.197
90 °C1.275

A 4 mm² copper conductor is 4.31 Ω/km at 20 °C and 5.16 Ω/km at 70 °C — a 20% larger voltage drop for the same current, from temperature alone.

Percentage Drop and Receiving-End Voltage

$$\Delta V\% = \frac{\Delta V}{V} \times 100 \qquad V_{receiving} = V_{sending} - \Delta V$$

The percentage matters more than the volts. Losing 5 V is trivial on a 1170 V string and catastrophic on a 48 V battery circuit.

Power Loss: I²R

$$P_{loss} = I^2 R_{loop}, \qquad R_{loop} = \frac{2 R_T L}{n_{parallel}}$$

$$\text{loss}\% = \frac{P_{loss}}{V \times I} \times 100$$

For a resistive DC circuit at a given current, loss% equals drop%. Use that as a check on your working.

Worked Example 1 — PV String

12 A, 600 V, 40 m one-way, 4 mm² copper at 70 °C.

  • R₇₀ = 5.157 Ω/km
  • R_loop = 2 × 5.157 × 0.040 = 0.4126 Ω
  • ΔV = 12 × 0.4126 = 4.95 V = 0.83%
  • Receiving-end voltage = 595.05 V
  • Loss = 12² × 0.4126 = 59.4 W = 0.83%

Comfortably inside a 1% target.

Worked Example 2 — Long String Run

14 A, 1170 V string, 100 m one-way, 4 mm² copper at 70 °C.

  • R_loop = 2 × 5.157 × 0.100 = 1.0314 Ω
  • ΔV = 14 × 1.0314 = 14.44 V = 1.23%
  • Loss = 14² × 1.0314 = 202 W = 1.23%

Still acceptable against a 2% limit. At 6 mm² the drop falls to 0.82% and the loss to 135 W — a genuine 67 W saving per string, every operating hour.

Worked Example 3 — Low-Voltage Battery Circuit

30 A, 48 V, 30 m one-way, 6 mm² copper at 70 °C.

  • R_loop = 0.2063 Ω
  • ΔV = 6.19 V = 12.9% — far beyond any acceptable limit

At 48 V this run needs roughly 50 mm² to stay under 2%. Low-voltage DC is dominated by voltage drop, never by ampacity. The dedicated 12V/24V/48V DC Cable Size Calculator is built for exactly this case.

What Changes the Drop

ChangeEffect on ΔV
Double the length×2
Double the current×2
Double the cross-section÷2
Copper → aluminium, same mm²×1.64
20 °C → 70 °C conductor×1.20
Two cables in parallel per polarity÷2

Acceptable Limits

CircuitCommon design limit
PV string / array DC1%
Array to inverter1–2%
Battery to inverter2%
General DC branch circuit3%
Absolute outer bound5%

These are design decisions taken in the project design basis, not universal code numbers. Tighter limits cost copper; looser limits cost energy every hour of operation.

Copper or Aluminium?

Aluminium needs roughly 64% more cross-sectional area for the same resistance. On short, high-current runs copper wins on space and termination simplicity. On long array-to-inverter feeders, aluminium at one or two sizes larger is often cheaper for the same electrical performance — provided the terminations are specified correctly.

Editorial standards

This guide is reviewed for formula, units and worked-example consistency. Standards and source organisations are named where they apply. Calculator results are educational aids and should be verified for your project, jurisdiction or personal circumstances.

K
Reviewed by Kodeeswaran Appavu
B.E. Civil Engineering graduate and solar design professional. Reviews OneCalcApp calculation guides for formula, units and practical assumptions.
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