Electrical Engineering · Worked example

How to Size a 48 V DC Feeder: Voltage Drop, Loss, and Current Density

A worked 48 V DC feeder example showing conductor resistance, voltage drop, I²R loss, current density, and why those checks do not replace code ampacity.

By Neutron Studios LLC8 minute readPublished 2026-08-11Reviewed 2026-08-11

Why this calculation matters

Low-voltage DC systems can carry large current even when the power level is modest. A conductor that looks acceptable by resistance alone can still waste meaningful power, run hot, or fail an installation-code requirement.

This example sizes a short 48 V copper feeder using the same screening quantities exposed by the Neutron wire-size and voltage-drop tools. The result is deliberately treated as an engineering screen, not as an NEC/CEC ampacity approval.

What you will calculate

Convert conductor resistivity and area into loop resistance.
Calculate voltage drop, delivered voltage, and cable loss.
Use current density as a thermal screening metric without confusing it with code ampacity.
Recognize when installation method, insulation rating, bundling, terminals, or protection rules require a separate code check.

Given values

DC bus voltage: 48 V
Load current: 40 A
One-way conductor length: 5 m
Copper conductor: 7 AWG, approximately 10.55 mm²
Conductor temperature used for resistance estimate: 30 °C
Copper resistivity at 20 °C: approximately 0.017241 Ω·mm²/m
Copper temperature coefficient: approximately 0.00393 /°C

Governing equations

Temperature-adjusted resistivity

\rho T = \rho _{20}\; \left[1 + \alpha \left(T - 20\; {}^{\circ} C\right)\right]

Copper resistance rises with conductor temperature.

DC loop resistance

R_{\mathrm{loop}} = \frac{\rho T\; \left(2 L\right)}{A}

A two-conductor DC circuit uses the outbound and return length.

Voltage drop

\Delta V = I\; R_{\mathrm{loop}}

For a DC feeder, the load voltage is approximately Vsource − ΔV.

Cable loss

P_{\mathrm{loss}} = I^{2}\; R_{\mathrm{loop}}

This is heat produced in the conductor pair.

Current density

J = \frac{I}{A}

Useful as a preliminary thermal screen, but not a regulatory ampacity calculation.

Worked solution

1. Adjust copper resistivity for temperature

At 30 °C, copper is about 10 °C above the 20 °C reference temperature. Applying the linear temperature coefficient gives a resistivity of about 0.01792 Ω·mm²/m.

\rho _{30} \approx 0.017241 \times \left[1 + 0.00393 \times 10\right] = \frac{0.01792\; \Omega \cdot \mathrm{mm}^{2}}{m}

2. Calculate the complete circuit resistance

The source-to-load distance is 5 m, but current must return to the source, so the resistive path is 10 m. Dividing by 10.55 mm² gives about 0.01699 Ω for the conductor pair.

R_{\mathrm{loop}} \approx \frac{0.01792 \times 10}{10.55} = 0.01699\; \Omega

3. Calculate drop and delivered voltage

At 40 A, the conductor pair drops about 0.679 V. Relative to a 48 V source, that is about 1.42%, leaving roughly 47.3 V at the load before other connector, fuse, contactor, or battery-internal drops are included.

\begin{gathered}\Delta V \approx 40 \times 0.01699 = 0.679\; V\\\text{drop} \approx 1.42 \%\\V_{\mathrm{load}} \approx 47.32\; V\end{gathered}

4. Calculate cable heating and current density

The same resistance dissipates about 27.2 W at 40 A. Current density is about 3.79 A/mm². Those numbers are useful for comparing conductor candidates, but neither establishes allowable ampacity for a real installation.

\begin{gathered}P_{\mathrm{loss}} \approx 40^{2} \times 0.01699 = 27.2\; \mathrm{W}\\J \approx \frac{40}{10.55} = \frac{3.79\; A}{\mathrm{mm}^{2}}\end{gathered}

Result

Engineering interpretation

For this 5 m, 40 A example, 7 AWG copper gives approximately 1.42% voltage drop and 27 W of conductor loss at the assumed 30 °C conductor temperature.

If the design target is 3% maximum drop, this conductor passes the voltage-drop screen with margin. The final conductor choice still needs the applicable ampacity, insulation, bundling, termination, protection, environment, and jurisdiction checks.

Sanity checks

Doubling current doubles voltage drop but quadruples I²R loss.
Doubling one-way length doubles both voltage drop and conductor loss.
A larger conductor area should reduce resistance, drop, loss, and current density.
If a result claims that a thinner wire has less resistance than a thicker wire of the same material and length, the inputs or units are wrong.

Common mistakes

Using one-way length instead of round-trip length for a two-conductor DC circuit.
Treating current density as NEC/CEC ampacity.
Ignoring resistance added by terminals, fuses, breakers, contactors, and connectors.
Using 20 °C resistance for a conductor expected to operate substantially hotter.

References and model boundaries

Copper resistivity and temperature-coefficient values are standard room-temperature engineering reference values.
Final conductor ampacity and protection must be checked against the governing electrical code and the actual installation method.

For safety-critical, regulated, production, or otherwise consequential work, independently verify the result using the governing standard, current manufacturer data, and qualified engineering review. See the site methodology and engineering disclaimer.