Choosing the right wire cross-section for a circuit is not only about current carrying capacity. A cable that can safely carry the load current can still cause problems if the run is long — because resistance builds up over distance, and voltage is lost before it reaches the equipment. This lost voltage is called voltage drop, and ignoring it is one of the most common errors in electrical installation design.
This guide covers the complete voltage drop calculation process for both single-phase and three-phase circuits, with worked examples, conductor resistance tables, and IS 732 compliance limits.
Every conductor has electrical resistance. When current flows through that resistance, a voltage is dropped across the cable — meaning the voltage at the end of the run is lower than at the source. The larger the current and the longer the cable, the greater this voltage drop.
IS 732 (Code of Practice for Electrical Wiring Installations) sets the following permissible voltage drop limits in India:
Lighting circuits: maximum 3% voltage drop (≈ 6.9V for 230V supply)
Power circuits and combined circuits: maximum 5% voltage drop (≈ 11.5V for 230V, ≈ 20.75V for 415V)
Measured from the supply point to the most distant point of the circuit
Exceeding these limits causes practical problems: lights that flicker or appear dim, motors that run hot due to higher current draw at lower voltage, and electronic equipment that malfunctions or fails prematurely.
The fundamental formula uses conductor resistance per unit length, load current, and cable length:
For single-phase (or DC) circuits:
VD (volts) = 2 × I × R × L ÷ 1000
Where: I = load current in amperes
R = conductor resistance in Ω per km (from table below)
L = one-way cable length in metres
The factor of 2 accounts for both the live and neutral conductors
For three-phase circuits:
VD (volts) = √3 × I × R × L ÷ 1000
√3 ≈ 1.732
L = one-way cable length in metres
This gives the line-to-line voltage drop; divide by √3 for line-to-neutral
To express as a percentage: VD% = (VD ÷ Supply Voltage) × 100
Use these resistance values for design calculations. Values are at 70°C, which is the standard operating temperature for PVC insulated cables. Resistance increases with temperature — using 70°C values is conservative and IS 732 compliant.
1.5mm² copper: 14.5 Ω/km | aluminium: 23.5 Ω/km
2.5mm² copper: 8.71 Ω/km | aluminium: 14.1 Ω/km
4mm² copper: 5.45 Ω/km | aluminium: 8.84 Ω/km
6mm² copper: 3.63 Ω/km | aluminium: 5.89 Ω/km
10mm² copper: 2.16 Ω/km | aluminium: 3.51 Ω/km
16mm² copper: 1.35 Ω/km | aluminium: 2.19 Ω/km
25mm² copper: 0.868 Ω/km | aluminium: 1.41 Ω/km
Note: These are resistance-only values. For large cables carrying high currents at power frequency, the inductive reactance component should also be included — but for cables up to 25mm², resistance dominates and reactance can be safely ignored for most practical calculations.
Scenario: A 30-metre run from a distribution board to a lighting point. Load current: 8A. Supply: 230V single-phase. Cable: 1.5mm² copper (standard for lighting in India).
VD = 2 × 8 × 14.5 × 30 ÷ 1000
VD = 2 × 8 × 14.5 × 0.030 = 6.96V
VD% = (6.96 ÷ 230) × 100 = 3.0%
Result: Exactly at the IS 732 limit. Acceptable, but no margin.
If the run were extended to 40 metres with the same load, VD would be 9.28V (4.0%) — exceeding the 3% limit for lighting. In that case, upsize to 2.5mm²:
VD = 2 × 8 × 8.71 × 40 ÷ 1000 = 5.57V = 2.4% ✓ Compliant
Lesson: for lighting runs beyond 30–35m at 8–10A, 2.5mm² is required
Scenario: 1.5 tonne split AC on a 25-metre run from the meter board. Running current: 8A (approx), but check voltage drop at starting current peak too. Cable: 2.5mm² copper (standard for AC circuits). Supply: 230V.
VD = 2 × 8 × 8.71 × 25 ÷ 1000 = 3.48V = 1.5% ✓ Well within limit
At starting current (2.5× running = 20A): VD = 2 × 20 × 8.71 × 25 ÷ 1000 = 8.71V = 3.8%
Starting transient is brief — not a sustained voltage drop — acceptable
Conclusion: 2.5mm² copper is correct for this run
Scenario: 5.5kW three-phase induction motor. Full load current: 11A (from nameplate or calculation). Cable run: 60 metres from MCC to motor terminal box. Cable: 2.5mm² copper, three-phase 415V supply.
VD = 1.732 × 11 × 8.71 × 60 ÷ 1000
VD = 1.732 × 11 × 0.5226 = 9.96V
VD% = (9.96 ÷ 415) × 100 = 2.4% ✓ Within 5% limit for power circuits
At starting (6× FLC = 66A): VD = 1.732 × 66 × 8.71 × 60 ÷ 1000 = 59.8V = 14.4%
Starting drop is transient but may prevent motor from starting — consider upsizing
With 4mm² copper: starting VD = 1.732 × 66 × 5.45 × 60 ÷ 1000 = 37.4V = 9.0%
With 6mm² copper: starting VD = 1.732 × 66 × 3.63 × 60 ÷ 1000 = 24.9V = 6.0% — motor will start reliably
This example illustrates an important industrial design point: even if steady-state voltage drop is within limits, motor starting voltage drop can prevent successful motor starting or cause contactors to chatter. For long motor cable runs, use steady-state voltage drop as a starting point, then verify starting drop separately.
For rapid field estimates with 2.5mm² copper cable on single-phase 230V circuits:
Voltage drop per ampere per 100m of run ≈ 1.74V (one-way) or 3.48V (loop)
Quick check: (Amps × metres) ÷ 144 = % voltage drop (approximate, 2.5mm² copper)
If result > 3 for lighting or > 5 for power: upsize the cable
For 4mm² copper, divide by 230 instead of 144 for the same quick check
The calculations above are sufficient for standard residential and light commercial wiring. For industrial installations, high-rise buildings, long feeder runs, or any installation where the fault current or load profile is uncertain, a qualified electrical engineer should perform the full design calculation — including reactive voltage drop components, harmonic effects, and protection coordination.
Voltage drop calculation is just one check. Always also verify that the selected cable meets the current-carrying capacity requirement (after all derating factors) and the short-circuit withstand requirement for the installation's fault level.
