Voltage Drop Explained: Cables, Resistance and Reactance

START WITH THE IDEA

Imagine a water pipe. When water flows, friction causes a pressure difference along the pipe. When electric current flows through a real cable, its electrical impedance causes a voltage difference between the two ends.

This lesson explains the basic physics, how to calculate a practical approximation, and what still needs checking before selecting a cable.

Open calculator →11 sections · diagrams · worked example

01 · The basic concept

What is voltage drop?

The voltage at the beginning of a cable is called the sending-end voltage. The voltage at the appliance, motor or distribution board is called the receiving-end voltage. Usually, when current flows, the receiving voltage is slightly lower.

What happens along a cable?Illustrative circuit

The cable is not an ideal wire. When current flows, the conductor impedance causes a voltage difference between the supply and the load. Diagram is conceptual, not a wiring or protective-earth drawing.

\Delta U\approx U_{\mathrm{sending}}-U_{\mathrm{receiving}}

We normally express voltage drop in volts (V) and as a percentage (%) of nominal supply voltage. The calculator estimates the in-phase voltage magnitude difference for one feeder under the stated assumptions.

Voltage along the feederIllustrative steady current

With constant current and uniform cable parameters, the approximate voltage drop accumulates along cable length. The graph is not to scale; real feeder voltage depends on the load and network.

02 · Practical engineering

Why should we care?

Every real conductor has impedance. If the voltage reaching a load is too low, equipment may malfunction, motors may deliver less starting torque, and equipment performance may suffer. Excessive conductor resistance also produces heating losses, but voltage drop and cable heating are separate design checks.

Supply quality

At the load

Equipment needs voltage within a suitable operating range.

Load current

More current, more drop

A heavily loaded cable generally has a larger voltage drop.

Cable route

Longer means more drop

The same conductor has more total impedance over a longer run.

Think of voltage drop as only one item on the cable-design checklist. The cable must also have sufficient ampacity, fault withstand and appropriate protection.

03 · Know your system

Why three-phase, single-phase and DC differ

Our calculator supports three common arrangements. Choosing the right arrangement matters because each uses a different voltage reference and conductor-path multiplier.

Balanced three-phase AC

Use line-to-line voltage (L-L), normally 400 V in a European 230/400 V system. For a balanced load, the line-to-line drop approximation includes a √3 multiplier.

\Delta U_{3\phi}\approx\sqrt{3}\times I\times L\times(R\cos\varphi+X\sin\varphi)

Single-phase AC, two wires

For a typical line-to-neutral circuit use line-to-neutral voltage (L-N), normally 230 V. Current travels to the load and returns through the neutral, so we use a factor of 2 when the conductors have equal impedance.

\Delta U_{1\phi}\approx2\times I\times L\times(R\cos\varphi+X\sin\varphi)

DC, two wires

There is no AC phase angle or inductive reactance term in this simple steady-state calculation. The outgoing and returning wires both contribute resistance.

\Delta U_{DC}=2\times I\times L\times R

Important: The factor 2 assumes two equal-conductor paths. If the neutral/return conductor is a different size, or if several loads share a feeder, the circuit must be modelled more carefully. A balanced three-phase circuit is not simply three separate single-phase calculations.

04 · Read the inputs

Current and length — the easy part

I is load current measured in amperes (A). L is the one-way route length from the supply to the load. In our equations resistance and reactance are in Ω/km, so we convert metres to kilometres.

L_{\mathrm{km}}=\frac{L_{\mathrm{m}}}{1000}

For example, a 50 m cable run means L = 0.05 km. Do not double that input: the single-phase and DC equations already account for the return conductor.

05 · Cable material

What is resistance, and why does temperature matter?

Resistance R describes how strongly a conductor opposes current. Copper typically has lower electrical resistivity than aluminium of the same cross-sectional area. A larger conductor gives current more cross-section to flow through, lowering resistance.

R_{20}=\frac{1000\times\rho_{20}}{A}

Here A is conductor area in mm², ρ₂₀ is reference resistivity in Ω·mm²/m, and the 1000 converts Ω/m to Ω/km. Real cables are made to a standard: IEC 60228 sets a maximum DC resistance for each size, which is typically 5–7 % higher than the ideal-metal value above. The calculator therefore offers a resistance source selector: the default pure-metal estimate, the IEC 60228 maximum (conservative, for sizes tabulated in the built-in data), or a manufacturer value. If a size is not tabulated, the calculator tells you it fell back to the resistivity estimate.

Conductors get more resistive as they warm up. The following linear correction is commonly used as a first approximation:

R_{\theta}=R_{20}\times[1+\alpha_{20}\times(\theta-20)]

Approximate copper ρ₂₀0.017241 Ω·mm²/mApproximate aluminium ρ₂₀0.028264 Ω·mm²/mCopper temperature coefficient α₂₀0.00393 / °CAluminium temperature coefficient α₂₀0.00403 / °C

Do not confuse estimates with cable data. The calculator’s automatic R uses simple resistivity values. Real manufactured cable resistance depends on conductor construction, tolerances and AC effects such as skin/proximity effect. For professional design, use verified manufacturer or appropriate standard conductor data at the operating temperature.

06 · Parallel cable runs

What changes when cables run in parallel?

Large feeders often need several conductors connected to the same phase. For example, 3 × 240 mm² copper per phase means three 240 mm² conductors share phase L1, three share L2 and three share L3. You do not enter 720 mm² as one conductor: select 240 mm² and 3 parallel runs.

Enter the total feeder current. If identical parallel paths have equal impedance, each carries the total current divided by the number of runs.

I_{\mathrm{run}}=\frac{I_{\mathrm{total}}}{n}

Example: 900 A in three parallel runs

I_{\mathrm{run}}=\frac{900}{3}=300\ \mathrm{A}

Under the same ideal equal-sharing assumption, the equivalent feeder resistance is the resistance of one conductor divided by three. In a simplified equivalent-reactance model we also divide X by three.

R_{\mathrm{eq}}=\frac{R_{\mathrm{run}}}{3}\qquad X_{\mathrm{eq}}\approx\frac{X_{\mathrm{run}}}{3}

The calculator uses equivalent R and X with total current, which is mathematically equivalent to using each run’s impedance with its share of current when the paths are identical.

Essential conditions: Run conductors should be of the same material and cross-section, have essentially the same length, compatible terminations and a layout that encourages equal current sharing. For single-phase and DC, the outgoing and return paths must be paralleled equivalently. Single-core AC cable geometry, magnetic coupling and asymmetrical placement can produce unequal currents; actual reactance may not simply equal X/n. Check cable ampacity, grouping derating, protective devices and short-circuit withstand independently.

For further reading, see the Schneider Electric guide to conductors in parallel ↗.

07 · Understanding AC

What is cable reactance, and where does cos φ fit?

With alternating current, a cable also has inductive reactance X, related to magnetic fields around conductors. Its value depends on geometry, spacing and system frequency. A common illustrative LV estimate is 0.08 Ω/km, but it is not accurate for every cable.

The load current may lag behind or lead the voltage. This is the phase angle φ, closely related to power factor cos φ. Both R and X contribute to the voltage-drop projection, but by different amounts.

Why power factor mattersConceptual vector projection

Resistance and reactance contribute differently. The drop formula projects their contributions using cos φ and sin φ. This conceptual diagram does not show the complete sending-end/receiving-end voltage phasors.

\sin\varphi=\sqrt{1-\cos^{2}\varphi}

R\cos\varphi\pm X\sin\varphi

Use + for a lagging inductive load and − for leading capacitive current in this approximate relationship. A sufficiently leading load can yield a negative calculated drop (voltage rise). That requires more detailed verification rather than treating it as an automatic design approval.

08 · Put it together

The voltage-drop equations

For one uniform AC feeder, calculate the conductor impedance contribution, multiply by load current and one-way length, then apply the circuit multiplier.

\Delta U_{3\phi}\approx\sqrt{3}\times I\times L\times(R\cos\varphi\pm X\sin\varphi)

\Delta U_{1\phi}\approx2\times I\times L\times(R\cos\varphi\pm X\sin\varphi)

\Delta U_{DC}=2\times I\times L\times R

From volts to percentage

\Delta U_{\%}=\frac{\Delta U}{U_{\mathrm{supply}}}\times100

For three-phase use L-L nominal supply voltage; for the typical single-phase circuit use L-N voltage; for DC use the DC supply voltage.

These are approximations. They are commonly used for preliminary steady-state AC voltage-drop calculation. Exact phasor modelling, voltage-dependent load current, multiple sections and unbalanced circuits can require a network-level calculation.

09 · Worked example

Let's calculate a 400 V feeder

EXAMPLE · 3-PHASE LV CIRCUIT

32 A over 50 m, copper 10 mm²

Assume 400 V L-L, conductor temperature 70 °C, cos φ = 0.90 (lagging), and X = 0.08 Ω/km.

1. Reference conductor resistance

R_{20}=\frac{1000\times0.017241}{10}=1.7241\ \Omega/\mathrm{km}

2. Adjust to 70 °C

R_{70}=1.7241\times[1+0.00393\times50]

R_{70}\approx2.0629\ \Omega/\mathrm{km}

3. Length and power factor

L=\frac{50}{1000}=0.05\ \mathrm{km}

\sin\varphi=\sqrt{1-0.9^2}\approx0.4359

4. Impedance contribution

Z_{\mathrm{proj}}=R\times0.9+X\times0.4359

Z_{\mathrm{proj}}\approx1.8915\ \Omega/\mathrm{km}

5. Voltage drop

\Delta U\approx\sqrt{3}\times32\times0.05\times Z_{\mathrm{proj}}

\boxed{\Delta U\approx5.24\ \mathrm{V}}

6. Percentage and receiving voltage

\Delta U_{\%}\approx\frac{5.242}{400}\times100\approx1.31\%

U_r\approx400-5.242=394.76\ \mathrm{V}

Results are calculated from full-precision inputs. Intermediate displayed numbers are rounded.

Change these values in the calculator →

10 · Design decisions

How much voltage drop is acceptable?

How long can this cable be?

Voltage drop grows in direct proportion to length, so the calculator can also tell you the longest one-way length at which this cable just reaches your drop limit. It rearranges the same equation, where the coefficient k is √3 for three-phase and 2 for single-phase or DC:

L_{max}=\frac{\Delta U_{limit}}{k\,I\,(R\cos\varphi+X\sin\varphi)}\qquad \Delta U_{limit}=\frac{\text{limit}\,\%}{100}\,U

For the example circuit above (32 A, 50 m, 10 mm² copper, 5 % limit), the longest length is about 191 m. This number answers one question only: how far the cable can go before voltage drop alone is exceeded. The cable must still pass its ampacity and short-circuit checks, which is what LV Cable Sizing is for. If the load is leading the reactive term can cancel the resistive drop, and then no length reaches the limit; the calculator says so instead of printing a number.

There is no single percentage appropriate for every project. The permissible drop depends on the supply arrangement, circuit use, national adoption of standards and any more demanding equipment specification.

As background, the informative guidance associated with IEC 60364-5-52 is commonly presented with examples such as 3% for lighting and 5% for other uses on installations supplied directly from a public LV supply. Other categories have different recommendations and conditions. These are not universal approval thresholds for every feeder.

KiloVolt therefore lets you enter your own voltage-drop target. The on-screen check compares the calculated percentage with that value; it does not certify compliance with any particular installation standard.

Design practice: Check voltage drop from the applicable installation origin to the end load. A local branch may appear acceptable on its own while the combined upstream and downstream drop is excessive. Motor starting, harmonics, unbalance and other transient effects require separate evaluation.

Authoritative references

IEC 60364-5-52:2009+A1:2024 — Wiring systems (IEC webstore) ↗

Schneider Electric Electrical Installation Guide — Voltage-drop calculation ↗

Schneider Electric Electrical Installation Guide — Voltage-drop guidance ↗

When the result is not feasible. The formulas are linear and assume a constant current. If the calculated drop reaches the supply voltage (100 %), the voltage at the load would be zero or negative. That is outside what this normal-load approximation can describe, although a real circuit can reach zero terminal voltage, for example in a fault, where the constant-current assumption no longer applies. KiloVolt then shows “Not feasible” instead of numbers, and still gives the maximum length for your target. Above 20 % the figures are shown with a caution, because the constant-current assumption is already unreliable; that threshold is a judgment, not a standard. In both cases the fix is a larger cross-section, more parallel runs, a shorter length or a lower current.

11 · Before you finish

Common mistakes to avoid

Entering the round-trip distance

Enter the one-way cable length. The factor 2 already accounts for a two-wire return circuit.

Using 230 V for the three-phase L-L calculation

Use the nominal reference that matches the formula: 400 V L-L for typical three-phase LV and 230 V L-N for line-neutral circuits.

Assuming cable resistance is fixed at 20 °C

Resistance increases with conductor temperature. A warm cable usually has a larger drop.

Assuming 0.08 Ω/km is always the correct X

Reactance varies with cable configuration, spacing and frequency. Replace illustrative values with verified cable data when needed.

Ignoring parallel-cable geometry and grouping

Additional runs lower ideal resistance but do not guarantee equal current sharing or enough ampacity. The layout, terminations and grouping factors still matter.

Treating voltage-drop percentage as a complete cable check

Still verify installation ampacity, protection, thermal short-circuit withstand, disconnection requirements and any applicable national rules.

Assuming every load is balanced and sinusoidal

Three-phase imbalance, harmonics, distributed loads and motor starting can require more detailed calculations.

PUT IT INTO PRACTICE

Try your own cable and load

Change voltage, current, cable length, conductor material and temperature, then see every step of the result.

Open voltage drop calculator →

Educational scope: This guide describes a preliminary steady-state single-feeder method. It does not provide certified IEC 60228 values, final cable selection, or a legal compliance determination.