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Formula

Show calculation steps (2)
  1. Voltage Drop (%)

    Voltage Drop (%): 3 Phase Voltage Drop Calculator

    Voltage drop as a percentage of the line-to-line voltage

  2. Voltage at Load

    Voltage at Load: 3 Phase Voltage Drop Calculator

    Line-to-line voltage minus the voltage drop

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Results

Voltage Drop
4.09
volts (V)
Voltage Drop (%) 1.023 %
Conductor Resistance 0.0875 Ω
Voltage at Load 395.91 V

What Is the 3 Phase Voltage Drop Calculator?

This tool estimates the voltage lost along a balanced three-phase cable run. Excessive voltage drop causes dim lighting, overheating, nuisance tripping and poor motor performance, so most installation standards limit it to around 3–5% of nominal voltage. The calculator works in metric units and applies to any balanced three-phase AC circuit.

The Formula

The line-to-line voltage drop is:

$$V_{drop} = \sqrt{3} \times I \times R \times \cos\phi$$

where \(I\) = load current in amperes, \(R\) = one-way conductor resistance in ohms, and \(\cos\phi\) = power factor. The conductor resistance is found from:

$$R = \frac{\rho \, L}{A}$$

with \(\rho\) = resistivity in \(\Omega\cdot\text{mm}^2/\text{m}\) (about 0.0175 for copper, 0.028 for aluminium), \(L\) = one-way length in metres, and \(A\) = cross-sectional area in \(\text{mm}^2\). The percentage drop is \(\frac{V_{drop}}{V_{LL}} \times 100\).

Three-phase circuit diagram showing source, three conductors and a load with voltage drop along the cable
Voltage drop occurs across each conductor between the source and the load in a three-phase system.

How to Use It

Enter the load current, the one-way cable length, the conductor material resistivity, the cross-sectional area, the power factor, and the line-to-line system voltage. The calculator returns the volts dropped, the percentage drop, the conductor resistance, and the resulting voltage at the load.

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Worked Example

A 30 A load is fed by a 50 m run of 10 mm² copper (\(\rho = 0.0175\)) at power factor 0.9 on a 400 V system. The resistance is:

$$R = \frac{0.0175 \times 50}{10} = 0.0875\,\Omega$$

The voltage drop is:

$$V_{drop} = \sqrt{3} \times 30 \times 0.0875 \times 0.9 \approx 4.09\,\text{V}$$

which is about \(1.02\%\) of 400 V — well within typical limits.

Diagram of a cable cross-section showing length, conductor area and current
Key inputs to the calculation: cable length, conductor cross-sectional area, current and power factor.

FAQ

Why the \(\sqrt{3}\) factor? In a balanced three-phase system the line-to-line drop relates to the per-phase drop by \(\sqrt{3}\), giving the familiar \(\sqrt{3} \, I R \cos\phi\) expression.

Should I include reactance? This simplified model uses only resistance, which is accurate for small cables. For large conductors and long runs, cable reactance \(X\sin\phi\) also contributes and increases the drop.

One-way or round-trip length? Enter the one-way length. The three-phase formula already accounts for the conductor geometry, so you do not double the distance.

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