What Is Voltage Drop and Why Does It Damage Equipment?

Imagine a long corridor packed with people trying to walk from one end to the other. The wider the corridor, the easier the flow. The longer it is, the more effort it takes. If the corridor is narrow and long, people arrive at the far end exhausted, having lost energy fighting through the crowd.

A cable works the same way. Its conductor — copper or aluminium — is not a perfect path for electrons. It has impedance: a combination of resistance (which converts electrical energy into heat) and reactance (which opposes changes in current). When current flows through this impedance, voltage is consumed along the cable’s length. The voltage that arrives at the equipment is always less than the voltage that left the switchboard.

This lost voltage is not recoverable. It is dissipated as heat in the conductor, warming the cable and its surroundings. The power wasted equals I²R — current squared times resistance. For a 100 A circuit on a cable with 0.05 Ω total resistance, that is 500 W of heat generated continuously inside your cable run.

The key formula is deceptively simple: ΔV = I × Z × L, where I is the current, Z is the impedance per metre, and L is the one-way cable length. For single-phase circuits, the factor of 2 accounts for the current flowing out on the active and returning on the neutral. For three-phase balanced circuits, the factor is √3.

Standards allow up to 5% voltage drop for power circuits. On a 400 V three-phase supply, 5% is 20 V lost in the cables. The motor receives 380 V instead of 400 V. That sounds close enough — until you understand the physics.

An induction motor’s torque is proportional to the square of the supply voltage. This is the critical relationship: T ∝ V². A 5% voltage reduction does not reduce torque by 5%. It reduces torque by approximately 9.75% (since 0.95² = 0.9025). A 10% voltage drop — which can happen during motor starting — reduces available torque by 19%.

For a motor that needs 90% of its rated torque just to keep a conveyor belt moving, a 10% voltage drop means the motor cannot maintain speed. It slows down, draws even more current (trying to compensate), heats up, and if the condition persists, the winding insulation degrades. Each 10°C rise above rated temperature halves the insulation life. A motor rated for 20 years of service can be destroyed in 2 years by chronic undervoltage.

Lighting is equally sensitive but in a different way. Incandescent lamps (still found in some industrial and hazardous area applications) produce light output roughly proportional to V^3.4. A 5% voltage reduction causes a 16% drop in light output — enough to take a workspace below minimum illumination requirements. LED drivers are more tolerant but will flicker or shut down entirely if voltage falls below their minimum input threshold, typically around 200 V for a 230 V rated driver.

Voltage drop is not a single number for a single cable. In a real installation, the total drop is the sum of every segment between the supply transformer and the final equipment:

1. Transformer regulation: The transformer itself drops voltage under load (typically 4–6% at full load for a distribution transformer). This is before the cables even begin.
2. Main feeder cable: From the transformer to the main switchboard. This cable carries the total building load and may run 20–50 metres or more.
3. Sub-main cable: From the main switchboard to a distribution board on a different floor or area. Another 30–100 metres in a large building.
4. Final subcircuit cable: From the distribution board to the actual equipment. This is often the longest run in percentage terms because the cable is the smallest.

Consider a real scenario: a 1000 kVA transformer at 4.5% regulation under 80% load gives 3.6% drop before the cables. The main feeder adds 0.8%. The sub-main adds 1.2%. The final subcircuit adds 1.5%. The total drop at the equipment is 3.6% + 0.8% + 1.2% + 1.5% = 7.1% — well beyond the 5% limit most standards impose, even though no single cable exceeded its allowable drop.

This is why experienced engineers budget the total allowable voltage drop across all segments. A common rule of thumb: allocate 1.5% for the main feeder, 1.5% for the sub-main, and 2% for the final subcircuit, totalling 5%. If transformer regulation is significant, the cable budget must shrink accordingly.

The consequences of excessive voltage drop are not theoretical. Here are documented failure modes:

• Motor overheating and failure: A 22 kW pump motor on a 150-metre cable run received only 365 V instead of 400 V (8.75% drop). The motor drew 15% more current than rated to maintain output. Within 18 months, the winding insulation failed and the motor burned out. Replacement cost plus production downtime exceeded the price of the correct cable by a factor of ten.
• Lamp flicker and complaints: An office lighting circuit at the end of a 60-metre run from the distribution board showed visible flicker whenever the air conditioning compressor on an adjacent circuit started. The starting current of the compressor caused a transient voltage dip of 8–12% at the lighting circuit, well below the 97% threshold where fluorescent lamps exhibit perceptible flicker. The fix required a dedicated sub-main to the lighting distribution board.
• Variable speed drive tripping: A VSD controlling a cooling tower fan tripped on undervoltage every afternoon when the building load peaked. The cumulative voltage drop from the transformer through two levels of distribution reached 11% during peak demand. The VSD’s undervoltage protection threshold was 85% of nominal — the voltage was dipping to 356 V (89% of 400 V), but the VSD’s internal DC bus voltage was further reduced by the rectifier, triggering the trip.
• Control relay dropout: A contactor coil rated for 400 V has a minimum holding voltage of approximately 340 V (85%). During a motor start on the same distribution board, the voltage dipped to 350 V. The contactor held — barely. But the 24 V DC control relay powered from a small transformer on the same supply dropped out at 20 V (83% of nominal), tripping the entire process line.

Every transformer has an impedance, expressed as a percentage on its nameplate (typically marked as u_k% or Z%). A transformer with 5% impedance means that at full rated load, the output voltage drops by 5% compared to the no-load voltage. At half load, the drop is approximately 2.5% (it scales roughly linearly with load for resistive loads).

This is often overlooked in voltage drop calculations. Engineers carefully calculate cable voltage drop but forget that the transformer has already consumed 3–5% of the voltage budget before the first metre of cable.

The relationship between no-load voltage and full-load voltage is: V_FL = V_NL × (1 − u_k% × load fraction / 100). For a transformer with a 415 V no-load output and 5% impedance at 80% load: V_FL = 415 × (1 − 0.05 × 0.8) = 415 × 0.96 = 398.4 V. You have already lost 16.6 V before any cable is involved.

Transformer tap changers can partially compensate. Most distribution transformers have off-load tap changers with ±2.5% or ±5% adjustment. Setting the tap to the +2.5% position raises the no-load voltage to approximately 425 V, which at 80% load and 5% impedance gives 408 V — much closer to the 400 V nominal. This is a free improvement that costs nothing but requires the transformer to be de-energised to change the tap position.

When a voltage drop calculation shows the result exceeding allowable limits, there are several engineering responses, listed from most common to least:

1. Increase cable cross-section: The simplest fix. Going from 16 mm² to 25 mm² reduces voltage drop by approximately 36%. The trade-off is cost and physical size — larger cables need larger containment.
2. Relocate the distribution board: Moving the DB closer to the load reduces cable length and therefore voltage drop proportionally. In new designs, the DB location should be optimised for minimum total cable length to all loads.
3. Use parallel cables: Two cables in parallel have half the impedance of a single cable. This is common for large feeder circuits where a single cable would be impractically large.
4. Raise the distribution voltage: For industrial sites, distributing at 690 V or even 3.3 kV to remote motor control centres dramatically reduces voltage drop. The higher voltage means lower current for the same power, and voltage drop is proportional to current.
5. Adjust transformer taps: As described above, setting the tap higher compensates for downstream voltage drop. This is a zero-cost solution if the transformer has tap facilities.
6. Improve power factor: For highly inductive loads, the reactive component of voltage drop can be significant. Power factor correction capacitors at the load reduce both the resistive and reactive voltage drop components.

Understanding voltage drop conceptually is the first step. The next step is calculating it for real circuits with actual cable data, installation methods, and operating temperatures.

The ECalPro Voltage Drop Calculator uses tabulated mV/A/m values from four international standards (AS/NZS 3008, BS 7671, IEC 60364, and the NEC) to compute voltage drop for any combination of cable size, length, current, power factor, and temperature. It shows the drop in volts, as a percentage, and compares the result against the applicable standard’s limits with clear pass/fail indicators.

For a deeper dive into the engineering methodology, including the full impedance formula and how mV/A/m values are derived from R and X, read the Voltage Drop Calculation Guide in the documentation section.