Voltage Drop Calculator

A cable can be thermally adequate and still deliver unusable voltage at the far end. On long site runs, voltage drop — not current capacity — usually decides the cable size.

How this calculation works

Every conductor has resistance. Current through resistance produces a voltage drop, so the voltage at the load is always lower than at the source. Over a short run it’s negligible. Over 200 metres of site cable feeding a motor, it isn’t.

Three phase: V_drop = √3 × I × L × (R·cosφ + X·sinφ) ÷ 1000 Single phase: V_drop = 2 × I × L × (R·cosφ + X·sinφ) ÷ 1000

Where L is the one-way run length in metres, R and X are the conductor resistance and reactance in ohms per kilometre, and cosφ is the power factor.

Note the difference: single phase uses a factor of 2 because current flows out and back along two conductors. Three phase uses √3.

What limit applies. [VERIFY — confirm current UAE / DEWA / ADDC design limits before publishing] Design practice commonly limits total voltage drop to around 4% from the origin of the installation for power circuits, with a tighter figure often applied to lighting. Confirm the requirement that applies to your project with the local authority and the project specification — these limits vary and are periodically revised.

Why it matters beyond compliance. Low voltage at a motor terminal means low torque — torque falls with the square of voltage, so a 10% voltage drop costs about 19% of the motor’s torque. The motor draws more current to compensate, runs hotter, and its life shortens. Undervoltage is a slow, expensive failure mode.

Worked example

A 30 kW motor 180 metres from the generator, three phase 400 V, 0.85 power factor, 50 mm² copper cable.

Design current: (30 ÷ 0.9 × 1000) ÷ (400 × 1.732 × 0.85) = 56.6 A.

For 50 mm² copper, resistance is roughly 0.387 Ω/km and reactance roughly 0.082 Ω/km at operating temperature.

V_drop = 1.732 × 56.6 × 180 × (0.387 × 0.85 + 0.082 × 0.527) ÷ 1000 = 1.732 × 56.6 × 180 × 0.372 ÷ 1000 = 6.6 V

As a percentage: 6.6 ÷ 400 = 1.65%. Comfortably within a 4% limit.

Now run the same load at 400 metres and the drop reaches roughly 3.7% — right at the limit, and that’s before you account for starting current, where the momentary drop is several times worse.

Common mistakes

Using the round-trip length. The formulas already account for the return path. Enter the one-way run.

Ignoring reactance. Fine on small cables, significant above about 35 mm² where reactance becomes a meaningful part of the impedance.

Checking only the running condition. Starting current is many times running current, so the momentary drop is much deeper. On a long run feeding a motor, check both.

Using resistance at 20°C. Conductor resistance rises with temperature. At operating temperature it’s roughly 20% higher than the cold figure.

FAQ

What is the maximum allowable voltage drop? Around 4% from the installation origin is common design practice for power circuits, with lighting often held tighter. Confirm the specific requirement with your project specification and the local supply authority.

Does voltage drop matter on a generator supply? More than on a mains supply, because the generator’s own voltage regulation adds to the cable drop. Both eat into the same margin.

How do I reduce voltage drop? Increase the conductor size, shorten the run, split the load across parallel cables, or move the distribution point closer to the load. On generator supplies, relocating the set is often the cheapest fix.

Does voltage drop cause overheating? Not directly in the cable — but the low voltage at a motor makes it draw more current, which heats the motor and the cable both.