Load Drop Calculator: How to Estimate Voltage Drop

A load drop calculator estimates the voltage drop along an electrical conductor given the load current, wire length, and conductor size. You enter the system voltage, current, cable length, and material, and the calculator returns the voltage drop in volts and as a percentage of the system voltage. This helps you size wires correctly so connected equipment receives adequate voltage.

Formula

The voltage drop formula used by this calculator is the standard DC approximation, which also works for single-phase AC circuits at power frequencies when the power factor is near 1:

V_drop = (2 × L × I × �) / A

• V_drop – voltage drop in volts (V)
• L – one-way length of the conductor in metres (m)
• I – load current in amperes (A)
• Ï� – resistivity of the conductor material in ohm-metres (Ω·m). For copper, Ï� ≈ 0.0175 Ω·mm²/m; for aluminium, Ï� ≈ 0.0282 Ω·mm²/m
• A – cross-sectional area of the conductor in square millimetres (mm²)

The factor 2 accounts for the current travelling out on the live conductor and returning on the neutral (or second phase conductor) in a single-phase circuit. For a three-phase balanced system, the voltage drop is typically calculated as √3 × (L × I × �) / A, but the single-phase formula here is a conservative approximation that is commonly used for general checks.

Worked Example

Using the default values from the calculator: system voltage 230 V, load current 10 A, one-way cable length 20 m, conductor cross-section 2.5 mm², copper material (� = 0.0175 Ω·mm²/m).

1. Calculate the round-trip length: 2 × 20 m = 40 m
2. Multiply by current and resistivity: 40 × 10 × 0.0175 = 7.0
3. Divide by the cross-section: 7.0 / 2.5 = 2.8 V
4. As a percentage: (2.8 / 230) × 100 = 1.22%

The calculator shows a voltage drop of 2.80 V and 1.22%, matching the manual calculation.

When to Use a Load Drop Calculator

Use this calculator when designing or troubleshooting any electrical installation where the distance between the source and the load is significant. Common situations include:

• Sizing cables for lighting circuits, socket outlets, or fixed appliances
• Checking whether an existing cable run is adequate for a new load
• Designing solar panel wiring, where long DC runs are common
• Planning low-voltage landscape lighting or security camera power feeds
• Verifying that voltage drop stays within recommended limits for sensitive equipment

Voltage drop matters because excessive drop means the load receives less voltage than it needs. Motors may overheat, lights dim, and electronic devices may malfunction or shut down. Most electrical standards recommend keeping voltage drop below 3% for branch circuits and 5% for the entire installation, but local codes may have specific values.

How to Read the Result

The output gives two numbers: the absolute voltage drop in volts and the percentage relative to the system voltage. A lower percentage is better. If the result exceeds your acceptable threshold (often 3% or 5%), you have three options:

• Increase the conductor cross-section (use a thicker wire)
• Shorten the cable run (move the source closer to the load)
• Reduce the load current (split the circuit or use more efficient equipment)

Note that the percentage is always relative to the nominal system voltage. A 5 V drop on a 12 V system is 41.7%, which is catastrophic, while the same 5 V drop on a 230 V system is only 2.2%, which is acceptable.

Common Mistakes and Misconceptions

Several errors frequently lead to incorrect voltage drop calculations:

• Forgetting the round-trip factor – Many people use the one-way length without multiplying by 2. This gives a result that is half the actual drop.
• Mixing units – Resistivity is often given in Ω·mm²/m, but if you use metres for length and square millimetres for area, you must keep them consistent. Using feet or circular mils without conversion will produce nonsense.
• Using the wrong resistivity – Aluminium has a higher resistivity than copper, so a copper wire of the same size will have a lower drop. Some tables give conductivity (the inverse of resistivity), which can be confusing.
• Ignoring temperature – Resistivity increases with temperature. The values used here are typical at 20°C. At higher operating temperatures, the actual drop will be slightly larger.
• Assuming the DC formula works for all AC cases – For AC circuits with reactive loads (motors, transformers), the impedance (not just resistance) matters. The DC formula is a good approximation for resistive loads and for short runs, but for long runs with inductive loads, a more detailed calculation is needed.

Limitations of the Simple Formula

The formula used here is a simplification. It assumes a purely resistive conductor and ignores:

• Reactance – In AC circuits, conductors have inductance and capacitance. At 50/60 Hz and for typical sizes, reactance is small compared to resistance, but it becomes significant for large conductors (above 25 mm²) or very long runs.
• Power factor – The formula assumes a power factor of 1 (pure resistive load). For inductive loads, the voltage drop is higher than the resistive-only calculation.
• Skin effect – At higher frequencies, current tends to flow near the surface of the conductor, increasing effective resistance. This is negligible at mains frequencies for normal sizes.
• Temperature rise – Under load, the conductor heats up, increasing its resistance. The calculator uses the 20°C value, so the actual drop in a hot cable will be higher.

For most residential and light commercial installations, the simple formula is accurate enough. For industrial installations with large conductors or long feeders, consult a qualified electrical engineer or use a more comprehensive calculation method that includes reactance.

Choosing the Right Cable Size

To use the calculator for cable sizing, you need to know the load current and the acceptable voltage drop. The procedure is:

1. Determine the load current (from equipment nameplates or design specifications).
2. Decide the maximum acceptable voltage drop (e.g., 3% of system voltage).
3. Enter your system voltage, current, length, and try different cross-sections until the percentage drop is below your limit.
4. Also check that the chosen cable can handle the current without overheating (ampacity). The voltage drop calculation does not replace the ampacity check.

For example, if you have a 20 A load at 230 V, 30 m away, and you want less than 3% drop (6.9 V), you can try a 4 mm² copper wire: drop = (2 × 30 × 20 × 0.0175) / 4 = 5.25 V (2.28%). That works. A 2.5 mm² wire would give 8.4 V (3.65%), which exceeds the limit.

Frequently Asked Questions

What is the difference between voltage drop and load drop?

“Load drop� is a colloquial term that usually refers to voltage drop under load. When a circuit is loaded, the current causes a voltage drop along the conductors, so the voltage at the load is lower than at the source. Some people also use “load drop� to mean the reduction in voltage when a large load is switched on, which is the same phenomenon.

Does the calculator work for DC and AC?

Yes, the formula is identical for DC and for single-phase AC at power frequencies when the power factor is close to 1. For three-phase AC, the voltage drop is slightly lower (multiply by √3/2 instead of 1), but using the single-phase formula is a safe conservative estimate.

What is a typical acceptable voltage drop?

Many electrical standards recommend a maximum of 3% for branch circuits and 5% for the total from the service point to the load. However, some equipment has specific voltage tolerances. Always check the manufacturer’s requirements and local codes.

Can I use this calculator for long runs like 100 metres?

Yes, but for runs longer than about 50 metres, the simple resistive formula may underestimate the drop if the power factor is low or if the conductor is large. For such cases, consider using a more detailed calculation that includes reactance. The calculator gives a good first estimate.

Why does the calculator multiply by 2?

The factor of 2 accounts for the current travelling out to the load on the live wire and returning on the neutral or second phase wire. Both conductors have resistance, so both contribute to the voltage drop. For a three-phase system, the return path is shared, so the factor is different (√3 instead of 2).

What if I only know the load in watts, not amps?

You can convert watts to amps using the formula: I = P / V for a resistive load. For example, a 1000 W heater on a 230 V supply draws about 4.35 A. For inductive loads, use I = P / (V × power factor).

Related

• Palworld Save File Location & Backup Guide
• Nikon Free Online Course: What You Get
• Arrange Notification Center by Time in OS X

in v | System voltage | V | 230 | 0
in i | Load current | A | 10 | 0
in l | One-way cable length | m | 20 | 0
in a | Conductor cross-section | mm² | 2.5 | 0
sel m | Conductor material | Copper=0.0175, Aluminium=0.0282 | 0.0175
out vd | Voltage drop | V | (2 * l * i * m) / a | 2
out pct | Voltage drop | % | (vd / v) * 100 | 2