In short. Voltage drop is the difference between the voltage at the generator terminals and at the load, caused by current flowing through the resistance and reactance of the cable. For a balanced three-phase circuit it is √3 × I × L × (R cos φ + X sin φ). Indian practice keeps the total within 3 to 5 percent at the furthest load, and motor starting is checked separately because the drop rises with the current.
Why the load sees less than the generator makes
An AVR holds the alternator terminals within about 1 percent of the set voltage, but it knows nothing about the far end of the cable. Every conductor has resistance and every circuit has reactance, so current produces a voltage difference along the run, and a generator can be perfectly regulated at 415 V while a pump 200 metres away sees 390 V. Equipment is generally built for about plus or minus 10 percent at its terminals, and that margin is shared between supply tolerance, source regulation and the cable, which is why the cable gets only a slice. The design intent is explained in generator cable sizing, of which the voltage-drop check is one part.
The steady-state calculation
ΔV = √3 × I × L × (R cos φ + X sin φ) for three phase; ΔV = 2 × I × L × (R cos φ + X sin φ) for single phase
ΔV is the voltage drop in volts, I is the load current in amperes, L is the one-way route length in kilometres, R and X are the conductor resistance and reactance in ohms per kilometre at the operating temperature, cos φ is the load power factor and sin φ follows from it. The result is divided by the nominal line-to-line voltage for three phase, or the line-to-neutral voltage for single phase, to give the percentage. Resistance is taken at the operating temperature, which for a fully loaded XLPE cable is 90 °C and raises R by about 28 percent over the 20 °C catalogue value. Length must be in kilometres when R and X are in ohms per kilometre.
Worked example (illustrative). A 415 V three-phase load of 120 A at 0.85 power factor is fed through 150 metres of 70 mm² copper XLPE cable. At 90 °C the resistance is about 0.34 Ω/km and the reactance about 0.075 Ω/km. With cos φ of 0.85 and sin φ of 0.53, the impedance term is 0.34 × 0.85 + 0.075 × 0.53, or 0.329 Ω/km. The drop is 1.732 × 120 × 0.15 × 0.329, or about 10.3 V, which is 2.5 percent of 415 V. That is acceptable for a feeder with a 3 percent allocation, but if the same cable also carries the start of a motor the next section applies.
| Copper XLPE multicore, mm² | Approximate drop, mV per A per m, three phase at 0.8 PF | Drop for 100 A over 100 m |
|---|---|---|
| 25 | 1.5 | 15 V, 3.6 percent |
| 50 | 0.81 | 8.1 V, 2.0 percent |
| 95 | 0.43 | 4.3 V, 1.0 percent |
| 150 | 0.29 | 2.9 V, 0.7 percent |
| 240 | 0.21 | 2.1 V, 0.5 percent |
The millivolt-per-ampere-metre values are typical of published tables and approximate; use the cable manufacturer’s figures for a design. The table shows the shape of the problem: beyond about 150 mm² reactance stops the drop falling in proportion to the extra copper, so long runs at large currents are better served by parallel cables, busduct or a higher voltage.
Limits and how to allocate them
Indian practice, following IS 732 and the National Building Code, generally keeps the drop from the origin of the installation to the furthest point within about 3 percent for lighting and 5 percent for other loads. The useful discipline is allocation: if a generator feeder, a sub-main and a final circuit are in series, each is given a share at the design stage, typically 1 to 2 percent for the feeder, so that later circuits are not forced into oversized cables. A generator also has excursions of 10 to 20 percent on load steps, so the cable allocation should not use up all the equipment margin. The relationship between steady-state regulation and transient behaviour is set out in generator transient response.
Motor starting is a different calculation
The formula does not change, but the current and power factor do. A direct-on-line motor draws six to eight times its rated current at a power factor of 0.2 to 0.3 for the first second or two, so a 2 percent run at full load becomes a 12 to 16 percent run during the start, with the weight of the drop shifting to the reactance term. Added to the generator’s own transient dip of 15 to 20 percent for a large motor, the motor terminals may see 65 to 70 percent voltage. Torque falls with the square of voltage, so starting torque is halved, a loaded pump or compressor stalls, contactors drop out below about 80 percent, and the overload trips. Reduced-current starters and a larger or shorter cable are traded against each other, as discussed in voltage dip and the motor-starting study guide.
Reducing voltage drop
- Increase the conductor size, which helps most below about 150 mm² where resistance dominates.
- Run two or more cables in parallel, which halves both resistance and reactance per circuit when laid in matched trefoil groups.
- Shorten the route or move the generator or the distribution board closer to the heavy load.
- Improve the load power factor with correction at the load, which reduces current for the same kW; capacitors must be checked for resonance on generator supply.
- Distribute at 11 kV for long distances or large campuses, with transformers near the loads, which cuts the percentage drop by the square of the voltage ratio.
Measuring and diagnosing drop on site
Suspected voltage-drop problems are confirmed by measuring at the source and the load at the same time under the same load. A single reading at the load says the voltage is low but not why; the generator terminal voltage, frequency and load current separate cable drop from AVR setting, unbalance or overload. Unbalanced single-phase loading shifts the neutral and appears as one phase high and another low at the far end. Long temporary runs on rental sites are a frequent source of low-voltage complaints, and the screening guide gives a quick check before the cable is laid.
Common mistakes
- Entering the route length in metres with ohms per kilometre; the drop comes out a thousand times too small and the error is not noticed until the pump will not start.
- Using the 20 °C resistance from the catalogue; the running cable is far hotter and the drop a quarter higher than calculated.
- Checking only the steady-state case; the motor start is the condition that fails.
- Applying the three-phase formula to a single-phase circuit; the round-trip conductor path doubles the drop.
What to send us
Send the circuit voltage and phase arrangement, the load current or kW and power factor, the one-way route length, the cable size, conductor material and construction if already installed, and the largest motor and its starting method. If the problem is on an existing site, add voltage readings taken at the generator and at the load at the same moment, with the load current. Send your application details and we will return the steady-state and starting figures with the options for correcting them.
Questions engineers ask
What is the acceptable voltage drop for a generator cable?
Common practice allows 3 to 5 percent from the source to the furthest load in total, with the generator feeder usually allocated 1 to 2 percent of that. Equipment with tight tolerances, such as UPS inputs and some drives, may need less, and the motor-starting case is checked separately.
Why is the voltage low at the load when the generator shows 415 V?
Because the drop occurs along the cable between the two points. Current through the cable’s resistance and reactance consumes part of the voltage, and the AVR cannot see or correct it. A longer run, a smaller cable, a heavier load or a poorer power factor all increase the loss.
Does voltage drop waste fuel?
The power lost in the cable is I² × R and is paid for in diesel, but at a few percent drop the energy cost is small. The larger costs are motors running hot at reduced voltage, equipment tripping, and a generator blamed for a cable problem.
How do I calculate single-phase voltage drop for a 230 V circuit?
Use ΔV = 2 × I × L × (R cos φ + X sin φ), with L as the one-way length, because the current travels out on the phase and back on the neutral. Divide the result by 230 V for the percentage. For the same current and cable, single-phase drop is about 15 percent higher than the three-phase line-to-line figure.
Technical content reviewed by Srinivas Reddy, Managing Partner, Reddy Generators. Last updated 19 October 2026.