Quick answer: Voltage drop is VD = (mV/A/m × I × L) ÷ 1000 volts, and the percentage drop is %VD = VD ÷ V × 100. Keep it within the 3% limit (IS 732) from the board to the load. If a run fails, the fix is almost always to increase the conductor size. Do it instantly with the voltage drop calculator.
The problem: it works on the bench, not at the end of the cable
Lights that dim when a motor starts, a contactor that chatters, a VFD tripping on under-voltage, an LED driver buzzing 80 m down a run — these are not usually faulty equipment. They are voltage drop. Every metre of cable has resistance, so the voltage at the far end is always lower than at the board. Push too much current down too long or too thin a cable and the load is starved.
The voltage drop formula
VD (V) = (mV/A/m × I × L) / 1000 I = design current (A) L = one-way cable length (m) mV/A/m = tabulated volt-drop of the cable %VD = VD / V_nominal × 100
The mV/A/m value already bundles the conductor's resistance and reactance, so you don't handle ohms directly. Use the single-phase (2-wire) figure below; for a balanced three-phase circuit, multiply it by 0.866 (√3/2).
Voltage drop table (copper, indicative mV/A/m)
| Conductor (mm²) | Single-phase mV/A/m | 3-phase (×0.866) |
|---|---|---|
| 1.5 | 29 | 25.1 |
| 2.5 | 18 | 15.6 |
| 4 | 11 | 9.5 |
| 6 | 7.3 | 6.3 |
| 10 | 4.4 | 3.8 |
| 16 | 2.8 | 2.4 |
| 25 | 1.75 | 1.5 |
| 35 | 1.25 | 1.08 |
| 50 | 0.93 | 0.81 |
| 70 | 0.63 | 0.55 |
| 95 | 0.47 | 0.41 |
Indicative values for 70 °C thermoplastic copper (aligns with BS 7671 Table 4D1B). Confirm against the cable maker's data for the final design.
The limit: 3% (and why)
IS 732 (and IEC 60364) target a maximum voltage drop of about 3% for final circuits, with roughly 5% allowed overall from the origin of the installation. Below that, motors start cleanly, lighting stays at rated lumens and control gear holds in. Above it, efficiency and equipment life suffer.
Worked example — does it pass?
A 32 A three-phase load, 415 V, on a 120 m run in 6 mm² copper:
mV/A/m (6 mm², 3-ph) = 6.3 VD = 6.3 × 32 × 120 / 1000 = 24.2 V %VD = 24.2 / 415 = 5.8% → FAILS the 3% limit
Now upsize the conductor and re-check:
10 mm²: 3.8 × 32 × 120 / 1000 = 14.6 V = 3.5% (still fails) 16 mm²: 2.4 × 32 × 120 / 1000 = 9.2 V = 2.2% ✓ PASS
So the run needs 16 mm², not 6 mm² — the current rating alone would have hidden this. That two-step check (ampacity and voltage drop) is exactly what the cable size calculator automates.
How to fix a failing voltage drop
- Increase the conductor size — the direct fix; each size up roughly cuts VD by a third.
- Shorten the run — relocate the sub-board or DB closer to the load so L drops.
- Raise the distribution voltage — feed at 415 V (3-phase) instead of 230 V where possible; %VD falls as voltage rises.
- Run cables in parallel — two 16 mm² in parallel behave like ~32 mm² for volt drop on very long feeders.
- Correct the power factor / reduce load — lower current means lower drop; capacitor correction helps on inductive loads.
FAQ
What is an acceptable voltage drop? 3% or less for a final circuit; up to about 5% total from the incomer per IS 732 / IEC 60364.
How do I calculate voltage drop quickly? Use VD = (mV/A/m × I × L)/1000, or enter current, length and cable size into the voltage drop calculator.
Is voltage drop the same as voltage loss? Yes — "voltage loss" and "volt drop" are the same thing: the fall in voltage along the cable due to its impedance.
Does length or current matter more? Both are linear in the formula, so doubling either doubles the drop — long runs to high-current loads are the usual culprits.